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IN   MEMORIAM 
FLOR1AN  CAJOR1 


A  SECOND  COURSE 
IN  ALGEBRA 


BY 


WEBSTER   WELLS,   S.B. 

PROFESSOR    OF    MATHEMATICS    IN    THE    MASSACHUSETTS 
INSTITUTE    OF    TECHNOLOGY 


BOSTON,   U.S.A. 
D.  C,  HEATH   &  CO.,  PUBLISHERS 

1909 


Copyright,  1909,  by  Webster  Wells 

All  rights  reserved 


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PREFACE 

In  the  preparation  of  this  text  the  author  acknowledges 
joint  authorship  with  Robert  L.  Short,  Technical  High  School, 
Cleveland. 

A  knowledge  of  the  more  elementary  parts  of  algebra  is 
presupposed.  For  this  reason  some  definitions  and  rules  for 
operation  are  assumed  as  already  known  to  the  pupil. 

Attention  is  called  to  the  generalization  and  bringing  together 
of  related  topics.  Chapter  III  is  an  example  of  this  feature. 
Here  all  forms  of  the  exponent  are  treated.  This  gives  oppor- 
tunity to  regard  the  logarithm  as  a  decimal  exponent  and  to 
make  the  logarithmic  operation  laws  intelligible.  The  intro- 
duction of  all  linear  equations  and  inequalities  in  Chapter  II 
shows  their  solution  directly  dependent  upon  the  four  funda- 
mental operations.  It  is  thought  that  the  introduction  of  the 
idea  of  functionality  and  of  algebraic  forms  taken  directly  from 
the  calculus  will  be  found  helpful  to  those  who  expect  to 
pursue  the  study  of  mathematics  further. 

The  treatment  of  factoring  is  thorough  and  so  taken  up  that 
Synthetic  Division  becomes  the  natural  method  for  factoring 
many  higher  forms  and  for  solving  equations  of  higher  degree. 

It  is  hoped  that  the  treatment  of  variation  as  a  proportion 
will  remove  the  reluctance  with  which  most  pupils  approach 
that  subject  in  connection  with  their  work  in  science. 

In  scope  this  text  is  sufficient  preparation  for  most  courses 

in   mathematics   which   require  thorough   knowledge   of   the 

operations  of  algebra. 

WEBSTER   WELLS. 


CONTENTS 

CHAPTER  PAGE 

I.     Fundamental  Laws  for  Addition  and  Multiplication  .  1 

II.     Addition,  Subtraction,  Multiplication,  Division  .         .  4 

Equivalent  Equations 11 

Equivalent  Systems  of  Equations 16 

Graphical  Kepresentation 21 

Inequalities 26 

III.  Exponents 32 

Miscellaneous  Examples 38 

Logarithms 41 

Properties  of  Logarithms .44 

Use  of  Table 48 

Applications 53 

IV.  Factors 57 

Miscellaneous  Examples 58 

Factor  Theorem  .........  60 

Horner's  Synthetic  Division 63 

Solutions  by  Factoring 65 

Common  Factors  and  Multiples 66 

V.     Fractions 73 

a    a 

0'a 76 

— ,   Ox  00,    00—  oo 80 

00 

Ratio  and  Proportion 82 

Variation      .         . .91 

VI.     Involution  and  Evolution  .         .         .         .         .         .97 

Series,  Binomial  Theorem 108 

Quadratic  Surds -  117 

VII.     Imaginary  Numbers     .         .         .         .         .         .         .         .  122 

Graphs  of  Imaginaries        .         .         .         ....  125 

VIII.     Quadratic  Equations  . 128 

Theory  of  Quadratic  Equations       .....  136 
v 


VI 


CONTENTS 


IX. 


X. 


XI. 

XII. 
XIII. 


XIV. 


XV. 


Graphs  of  Quadratic  Equations 

Discussion  of  Quadratic  Equations 

Problems  in  Physics 

Factoring 

Si  mi  ltaneous  Quadratic  Equations 

Graphs  ..... 

Series 

Arithmetic  Progressions    . 

Geometric  Progressions 

Infinite  Series     . 

convergency  and  divergency 

Summation     .... 

Differential  Method 

Interpolation 

Undetermined  Coefficients 

Partial  Fractions        .         .    , 

Permutations  and  Combination 

Determinants 

Properties  of  Determinants 

Evaluation   .... 

Solution  of  Equations 

Theory  of  Equations  . 

Transformation  of  Equations 

Descartes'  Rule  of  Signs 

Limits  of  Hoots   . 

Derivatives  .... 

Multiple  Roots   . 

Sturm's  Theorem 

Solution  of  Higher  Equations 

Reciprocal  or  Recurring  Equations 

Binomial  Equations 

Cubic  Equations  . 

Cardon's  Method 

Biquadratic  Equations 

EULBB'fl  Method  . 

Horner's  Method 


PAET   I 


ALGEBRA 


I.     THE  FUNDAMENTAL    LAWS   FOR  ADDITION  AND 
MULTIPLICATION 

1 .  The  Commutative  Law  for  Addition. 

If  a  man  gains  $8,  then  loses  $3,  then  gains  $6,  and  finally 
loses  $  2,  the  effect  on  his  property  will  be  the  same  in  what- 
ever order  the  transactions  occur. 

Then,  the  result  of  adding  +$8,  -$3,  +  $6,  and  -$  2, 
will  be  the  same  in  whatever  order  the  transactions  occur. 

Then,  omitting  reference  to  the  unit,  the  result  of  adding 
-f  8,  —  3,  +6,  and  —  2  will  be  the  same  in  whatever  order  the 
numbers  are  taken. 

This  is  the  Commutative  Law  for  Addition,  which  is : 

The  sum  of  any  set  of  numbers  will  be  the  same  in 
whatever  order  they  may  be  added. 

2.  The  Associative  Law  for  Addition. 

The  result  of  adding  b  +  c  to  a  is  expressed  a  -f-  (b  +  c), 
which  equals  (b  +  c)  +  a  by  the  Commutative  Law  for  Addi- 
tion (§  1). 

But  (b  -f-  c)  +  a  equals  6-f-c-f-a ;  and  b+ c-\-a  equals  a+b-\-c, 
by  the  Commutative  Law  for  Addition. 

Whence,  a  +  (b  +  c)  =  a  +  b  +  c. 

Then,  to  add  the  sum  of  a  set  of  numbers,  we  add  the 
numbers  separately. 

This  is  the  Associative  Law  for  Addition. 

3.  The  Commutative  Law  for  Multiplication. 

The  product  of  a  set  of  numbers  will  be  the  same  in 
whatever  order  they  may  be  multiplied. 

1 


2  ALGEBRA 

The  sign  of  the  product  of  any  number  of  terms  is  inde- 
pendent of  their  order ;  hence,  it  is  sufficient  to  prove  the 
commutative  law  for  arithmetical  numbers. 

Let  there  be,  in  the  figure,  a  stars  in  each  row,  and  a  in  a  row. 
b  rows.  ***  ••• 

We  may  find  the  entire  number  of  stars  by  multiply-  ***  ••• 

ing  the  number  in  each  row,   a,  by  the  number  of  ***... 

rows,  b.  

Thus,  the  entire  number  of  stars  is  a  x  b.  b  rows. 

We  may  also  find  the  entire  number  of  stars  by  multiplying  the  num- 
ber in  each  vertical  column,  &,  by  the  number  of  columns,  a. 
Thus,  the  entire  number  of  stars  is  b  x  a. 

Therefore,  a  x  b  =  b  x  a, 

which  is  the  law  for  the  product  of  two  positive  integers. 
Again,  let  c,  d,  e,  and  /  be  any  positive  integers. 

C  P         C  X  P  ' 

Then,    -  x-  -  = ;    for,    to    multiply   two    fractions,   we 

'    d     f     dxf  J 

multiply  the   numerators  together  for   the  numerator  of   the 

product,  and  the  denominators  together  for  its  denominator. 

O         P        P  X  c 

Then,   -x-== ;  since  the  commutative  law  for  multi- 

d     f    fxd 
plication  holds  for  the  product  of  two  positive  integers. 

Hence,   -  x  -  =  -  x  - ;  which   proves  the   commutative   law 

d     f     f     d1  * 

for  the  product  of  two  positive  fractions. 

4.   The  Associative  Law  for  Multiplication. 

To  multiply  by  the  product  of  a  set  of  numbers,  we 
multiply  by  the  numbers  of  the  set  separately. 

The  result  of  multiplying  a  by  be  is  expressed  a  X  (he), 
which  equals  (be)  x  a,  by  the  Commutative  Law  for  Multi- 
plication. 

(be)  x  a  equals  bca,  which  equals  abc  by  the  Commutative 
Law  for  Multiplication. 

Whence,  a  x  (6c)  =  abc 


FUNDAMENTAL   LAWS  3 

This  proves  the  law  for  the  product  of  three  numbers. 

The  Commutative  and  Associative  Laws  for  Multiplication  may  be 
proved  for  the  product  of  any  number  of  arithmetical  numbers. 
(See  the  author's  Advanced  Course  in  Algebra,  §§  18  and  19.) 

5.    The  Distributive  Law  for  Multiplication. 

The  law  is  expressed  (a  -f-  b)c  =  ac  +  be. 

We  will  now  prove  this  result  for  all  values  of  a,  b,  and  c. 

I.  Let  a  and  b  have  any  values,  and   let  c  be   a   positive 
integer. 

Then,  (a.  -f  b)c  =  (a  +  b)  +  («  +  &)  +  ...  to  c  terms 

=  (a  +  a+  •••  to  c  terms)  -J-  (p'+\b  +  •••  to  c  terms) 

(by  the  Commutative  and  Associative  Laws  for  Addition), 

=  ac  +  6c. 

II.  Let  a  and  6  have  any  values,  and  let  c  =  -,  where  e  and 
/are  positive  integers.  * 

Since  the  product  of  the  quotient  and  divisor  equals  the  dividend, 

Then,  (a  +  b)  x  -  x  /  =  (a  +  6)  x  e  =  ae  +  6e,  by  I. 

Whence,  (a  +  6)  x-x/=ax-x/+6x-x/. 

Dividing  each  term  by  /,  we  have 

(a+  &)  x  -=  a  x  -+6  x.-. 

^  /  /  / 

Thus,  the  result  is  proved  when  c  is  a  positive  integer  or  a 
positive  fraction. 

III.  Let  a  and  b  have  any  values,  and  let  c  =  —  #,  where  # 
is  a  positive  integer  or  fraction. 

(a  +  b)(-g)  =  -  (a  +  b)g  =  -  (ag  +  bg),  by  I  and  ft, 
=  -  ag  -bg  =  a(-  g)  +  6(-  gr). 

Thus,  the  distributive  law  is  proved  for  all  positive  or  nega- 
tive, integral  or  fractional,  values  of  a,  5,  and  c. 


4  ALGEBRA 

II.   ADDITION,    SUBTRACTION,    MULTIPLICATION,    DIVISION, 
APPLICATIONS 

6.  Similar  terms  are  those  which  do  not  differ  at  all  or 
differ  only  in  their  coefficients. 

7.  Any  factor  of  a  product  may  be  considered  the  coefficient 
of  the  product  of  the  remaining  factors. 

8.  To  add  two  similar  terms,  write  their  coefficients  with 
the  proper  sign  and  affix  the  common  literal  part. 

Ex.  1.     Find  the  sum  of  ax  and  bx. 

ax  +  bx  =  (a  +  b)x. 

Ex.  2.     Find  the  sum  of  3  abcx  and  —  5  mcx. 

3  abcx  +  (—  5  mcx)  =  (3ab  —  5  m)cx. 

This  is  equivalent  to  taking  the  common  factor  ex  from  the  expression 
3  abcx  —  5  mcx. 

9.  To  subtract  two  similar  terms  find  what  number  added 
to  the  subtrahend  will  produce  the  minuend.  The  number 
added  is  called  the  difference.  This  is  equivalent  to  chang- 
ing the  sign  of  the  subtrahend  and  adding  the  result  to  the 
minuend. 

Ex.  3.     Subtract  3  ax  from  5  ax.     2  ax  added  to  3  ax  is  5  ax. 
Hence  2  ax  is  the  difference. 

Ex.  4.     From  15  m  take  —8  m.     Changing  the  sign  (men- 
tally) of  —8  m,  we  have  15  m  4-  8  m  =  23  m. 
The  written  work  should  appear  in  this  form  : 
15  m 
—  8m 
23  m 

10.   Three  laws  enter  into  multiplication  of  monomials : 
TJie  law  of  shjiis. 
The.  law  of  coefficients. 
The  late  of ' exjumciitx. 


FUNDAMENTAL   PROCESSES  5 

The  product  of  two  terms  of  like  sign  is  positive ;  the 
product  of  two  terms  of  unlike  sign  is  negative. 

To  the  product  of  the  numerical  coefficients  annex  the 
letters ;  giving  to  each  an  exponent  equal  to  the  sum  of 
its  exponents  in  the  factors. 

The  same  three  laws  enter  into  division,  except  that  quotient 
takes  the  place  of  product  and  the  exponent  of  the  divisor  is 
subtracted  from  the  exponent  of  the  same  letter  in  the  divi- 
dend. (Make  a  rule  for  division  of  monomials.)  The  reason 
for  such  rule  follows  readily  when  division  is  defined  as  the 
process  of  finding  one  of  two  numbers  when  their  product  and 
one  of  the  numbers  are  given. 

11.  An  equation  is  a  statement  that  two  numbers  are  equal. 

12.  If  an  equation  is  true  for  all  finite  values  of  the  un- 
known numbers  involved,  it  is  an  identical  equation  or  identity. 

13.  If  an  equation  is  true  only  for  a  definite  set  of  values 
of  the  unknown  numbers  involved,  it  is  an  equation  of  condition. 

14.  An  equation  may  not  be  true  for  any  values  of  the  un- 
knowns involved.     It  is  then  said  to  have  no  roots. 

15.  If  when  a  number  is  substituted  for  an  unknown  in  an 
equation,  the  equation  becomes  identical  (§  12)  for  that  num- 
ber, the  equation  is  said  to  be  satisfied. 

The  roots  of  an  equation  are  the  numbers  which  satisfy  it. 
A  root  of  an  equation  is  also  called  a  solution  of  the  equation. 

16.  Some  principles  used  in  the  solution  of  equations  are  a 
set  of  generally  accepted  truths  called  axioms.  The  axioms 
most  frequently  in  use  are : 

1.  If  the  same  number,  or  equal  numbers,  be  added  to 
equal  numbers,  the  resulting  numbers  will  be  equal. 

2.  If  the  same  number,  or  equal  numbers,  be  sub- 
tracted from  equal  numbers,  the  resulting  numbers 
will  be  equal. 

3.  If  equal  numbers  be  multiplied  by  the  same  num- 
ber, or  equal  numbers,  the  resulting  numbers  will  be 
equal. 


6  ALGEBRA 

4.  If  equal  numbers  be  divided  by  the  same  number, 
or  equal  numbers  except  0,  the  resulting  numbers  will 
be  equal. 

17.  To  solve  an  equation  is  to  find  its  roots. 
The  following  steps  indicate  the  process  : 

$3-5  =  15.  (1) 

Add  5  to  each  member,  (Ax.  1) 

\x  =  15  +  5  =  20.  (2) 

Multiply  each  member  by  3,  (Ax.  3) 

2sc  =  60.  -(3) 

Divide  each  member  by  2,  (Ax.  4) 

x  -  30.  (4) 

18.  Two  equations  are  equivalent  when  every  solution  of  the 
one  is  a  solution  of  the  other. 

Thus  equations  (1),  (2),  (3),  (4)  are  equivalent. 

The  axioms  of  algebra  enable  us  to  transform  an  equation 
into  an  equivalent  one  which  may  be  more  easily  solved  than 
the  given  one. 

EXERCISE  l 
i.  Add  3a-  2  b  +  5c,  b-  9a- 11  c,  3c  +  b-  2  a,  b-  c-  a. 

2.  From  the  sum  of  7  x  —  8  ?/  -f-  4  z  and  -*2x  +  Bz+.f  take 
the  sum  of  x  —  y  —  z  and  y  +  z  —  9  x. 

3.  Add  3(?w  +  n)—  5s  +  £;     —  8(m  +  n)  +  4«  —  11  s; 

8  s  -  9  (m  +  n)  —  5 1 ;  6(m  +w)  —  4  g  +  9  t. 

4.  From   |p— fa-j-r  take  the  sum  of  |j>  +  ^ a  +  f  ?*  and 
\p-%q-\r. 

5.  Subtract  a#  4-  by  +  02;  from  m2#  —  y  +  <2& 

6.  Subtract  (c— d)»—  (c+d)y  from  (2e+6d)aM-(4c  —  3d)y, 

7.  Take  mv -f  a?  from  dkI  —  x2. 

8.  From    4aWc+5o6(c  +  d)  —  9a*6c?    take 

(3  a  +  5)  &*c  -  aft  (c  H 


FUNDAMENTAL  PROCESSES  7 

9.   Simp]  i  f'y    ( x*  —  4  x2  -f  5  x  —  1)  —  (2  ar3  +  5  a2  —  #  —  7)  + 
(or5  +  2  ar-  3  as  +  2). 

10.  Simplify  (x  + 1)  (x  -  2)  (a;  -  3)  -  (x  -  2)2  +  (ar3  -  1). 

11.  Simplify  (x  +  y)4  -  (x  -  y)\ 

12.  Simplify  [4  a;2  -(2  x  4-5)] [2  x2  -  (x  -3)]. 

13.  Multiply  4  a;2  +  ^  —  2/2  by  3  a?2  —  5  a;?/  -|-  4  1/2. 

14.  Multiply  a  a;  -|-  by  -f  cz  by  bx  —  ay  -\-  cz. 

15.  Multiply  4  (m  +  »)s  —  5  (wi  4-  w)  4-  7 

by  (m  4-  ^)2  +  2(m  +  n)-f  1. 

16.  Multiply  ar2a+1 4-  xayh  4-  y2h  by  a;a  —  yh. 

17.  Expand  (4  a  4-  3  &)2(4  a  -  3  b)2. 

18.  Multiply  1  a2  -  \  ab  +  f b*  by  - f a  4- 1 6. 

19.  Multiply  a2*  4-  <&'&«  4-  &2e  by  a2flr  -  agbe  4-  7re. 

»20.  Multiply  x2  —  #!/  4-  2/2  —  xz  —  yz  +  z2  by  a;  4-  y  4-  2. 

21.  Multiply  x2  +  ax-\-  bx  4-  a&  by  x  4-  c. 

22.  Divide  6x6~  19a;5  +  12 a?4 4- 5 ar3 4-4 a;2-  6x -2 

bv  2  a,*2 3  a; 1 

23.  Divide  a12  +  ft12  by  a4  4-  b\  J 

24.  Divide  32  m5  —  243  w3  by  2  m  —  3  n. 

25.  DivideTiFa3  +  ^^3byia4-|&. 

26.  Divide  a6n  —  66w  by  a2n  4-  a*&"  4-  b2n. 

27.  Divide  -i  a^  4-  37g  xl!  +  i  !/3  by  \  x  4- 1  y. 

28.  Divide  9  rV  +  15  r4  -  38  r8*  -  8  s4  -  26  r.s3  by  5  r8  4- 4  s2 -  ra 

29.  Divide  7  m2*+4  -  8  m'+V*-1  -  12  nAx~2  by  mx+2  -  2  n**~* 

30.  Divide  a;3  +(a  +  &) a;2 -  (6 a2  -  5 ab) x  +  6 a2b  by  a?  4- 3 a. 

Solve  the  following  equations  and  verify  results : 

31.  (x  +  2)(x-5)  =  x2-±x-4.. 


8  ALGEBRA 

32.   6(o;-3)+5(4flj-7)+l  =0. 

34.    2(3a:_2)_i(3^-2)  =  i(3^-2)-17. 
3S  (2/  -  4)  (2/  -f-  3)  (2/  -  2)  =  (2/  - 1 ) 3  -  1. 

Gt-\-T)     13     2*  ,   t 
6         15         21       5      3 

37.  ab  +  ax-\-  3b2  — 2 a2  =  4  be  —  bx-\-cx—c2—ac.    Solve  for  #. 

38.  y  —  e  = (x  —  d).     Solve  for  x. 

m 

39.  (a  +  b  4-  c)  (a?  —  2  a)  —  (a  —  c)  (a  -f-  &) 

=  (a_6_c)2_(a2  +  ^ 

ax—  b  ,  bx  —  c  ,  ex  —  a     A 

40. 1 - 1 =  0. 

a  b  e 

19.  It  is  sometimes  convenient  to  indicate  operations  of 
addition  and  subtraction.  For  this  purpose  parentheses  are 
used.  The  various  forms  of  parentheses  are  :.  parentheses  (), 
braces  \  \,  brackets  [  ],  and  the  vinculum  . 

A  positive  sign  before  parentheses  indicates  that  the  number 
within  is  to  be  added.  Hence,  parentheses  preceded  by  a  -f 
sign  may  be  removed  without  changing  the  signs  of  the  terms 
within. 

Ex.    2a  +  3&  +  (3a-5&  +  c)  =  2a  +  3&  +  3a-5&  +  c. 

A  negative  sign  before  parentheses  indicates  that  the  num- 
ber within  the  parentheses  is  to  be  subtracted.  Hence,  paren- 
theses preceded  by  a  —  sign  may  be  removed  if  the  -f-  signs  of 
the  terms  within  be  changed  to  —  and  the  —  signs  to  +  (§  9). 

Ex.  1.  5a  +  36 -(4  a  +  lb)  « 5a +3 b  - 4a  —  7 b  =  a- 4 6. 

Ex.  2.   5a+3  6-(-4a  +  7&)  =  r>a+3&+4a-7  6  =  9a--l/>. 

If  the  expression  contains  two  or  more  parentheses,  one 
within  the  other,  remove  one  at  a  time  beginning  with  the 
inner  parentheses. 


FUNDAMENTAL   PROCESSES  9 

Ex.  5  a  +  {3  a  -  (5  b  +  2  a)}  =  ■ 

6  a  +  {3  a  -  5  fr  -  2  a}  = 
5  a  +  3  a  —  66  —  2  a  = 
6  a  -  5  6. 

EXERCISE   2 

Simplify  the  following  by  removing  the  signs  of  aggregation, 
and  then  uniting  similar  terms  : 

i.  .9  m  +(—  4  m  +  6  n)  —  (3  m  —  n). 

2.  2x-3y-[ox  +  y]  +  \-Sx-7y\.     . 

3.  4?/-  2 a2 -[-4a;2  -  7  xy  +  5  y2~]  +  (8  x2  -  9  xy). 


4.  3a2-5ab-l-±a2  +  2ab-9b2l-7a2-6ab  +.62. 

5.  5  a  -(7  a  -[9  a  +  4]). 


6.  7x-\-8y-10x-lly\. 

7.  6  mn  -f-  5  —  ([  —  7  m/i  —  3]  —  |  —  5  mn  —  11  \ ). 

8.  2a-(-3  6+c-Ja-&j)-(3a  +  2c-[-2&  +  3c]> 


9.   37_[4i_{i3_(56-28  +  7)}]. 

10.  9  m  — (3  n  -f  J4  m  —  [n  —  6  m]  |  —  [m  -f  7  nj). 

11.  In  each  of  the  above  expressions  find  the  value  if 
a  ==  1,  b  ±=  —  2,  c  =  —  3,  m  =  5,  n  =  2,  a;  =  —  4,  ?/  =  —  1. 

20.  A  number  may  be  enclosed  in  parentheses  preceded  by 
a  +  sign  without  changing  the  sign  of  its  terms,  but  if  a  num- 
ber is  enclosed  in  parentheses  preceded  by  a  —  sign,  each  plus 
term  placed  in  parentheses  is  changed  to  minus  and  each  minus 
term  to  plus. 

EXERCISE   3 

In  each  of  the  following  expressions,  enclose  the  last  three 
erms  in  parentheses  preceded  by  a  —  sign : 

1.  a  —  b  *-  c  -f  d.  3.    x  -f-  x2y  —  xy2  —  y5. 

2.  ?ft3  +  2m2  +  3m  +  4.  4.    a2-4  62  +  12  6  -  9. 


10  ALGEBRA 

5.  4  x-  —  y2  —  2  yz  —  z2.  7.    x2  —  2  xy  +  y2  -f  3  x  —  1  //. 

6.  a2  +  //_c2  +  c^  8.    u4_5r*3-8n2  +  Gn  +  7. 

DEGREE   OP   A   RATIONAL   EXPRESSION 

21 .  A  monomial  is  said  to  be  rational  and  integral  when  it 
is  either  a  number  expressed  in  Arabic  numerals,  or  a  single 
letter  with  unity  for  its  exponent,  or  the  product  of  two  or 
more  such  numbers  or  letters. 

Thus,  3  a2bs,  being  equivalent  to  3  •  a  ■  a  •  b  •  b  •  ft,  is  rational  and  inte- 
gral. 

A  polynomial  is  said  to  be  rational  and  integral  when  each 

3 

term  is  rational  and  integral :  as  2  ar ab  -f-  c3. 

4 

22.  If  a  term  has  a  literal  portion  which  consists  of  a  single 
letter  with  unity  for  its  exponent,  the  term  is  said  to  be  of  the 
first  degree. 

Thus,  2  a  is  of  the  first  degree. 

The  degree  of  any  rational  and  integral  monomial  (§  21)  is 
the  number  of  terms  of  the  first  degree  which  are  multiplied 
together  to  form  its  literal  portion. 

Thus,  5  ab  is  of  the  second  degree;  3  a263,  being  equivalent  to 
3  •  a  •  a  •  b  •  b  •  ft,  is  of  the  fifth  degree  ;  etc. 

The  degree  of  a  rational  and  integral  monomial  equals  the 
sum  of  the  exponents  of  the  letters  involved  in  it. 

Thus,  rt64c3  is  of  the  eighth  degree. 

The  degree  of  a  rational  and   integral  polynomial   is  the 
degree  of  its  term  of  highest  degree. 
Thus,  2  a2b  —  3  c  +  d2  is  of  the  third  degree. 

23.  If  a  rational  and  integral  monomial  (§  21)  involves  a 
certain  letter,  its  degree  with  respect  to  it  is  denoted  by  its 
exponent. 


EQUIVALENT   EQUATIONS  11 

If  it  involves  two  letters,  its  degree  with  respect  to  them  is 
denoted  by  the  sum  of  their  exponents  ;  etc. 

Thus,  2  ab4x2ys  is  of  the  second  degree  with  respect  to  x  and  of  the 
fifth  with  respect  to  x  and  y. 

24.  An  Integral  Equation  is  one  each  of  whose  members  is 
a  rational  and  integral  expression  (§  21)  ;  as, 

4a?  —  5  =  \y  + 1. 

A  Numerical  Equation  is  one  in  which  all  the  known  num- 
bers are  represented  by  Arabic  numerals  ;  as, 

2  x  — 7  =  x  +  6. 

25.  If  an  integral  equation  (§  24)  contains  one  or  more  un- 
known numbers,  the  degree  of  the  equation  is  the  degree  of  its 
term  of  highest  degree. 

I     Thus,  if  x  and  y  represent  unknown  numbers, 
ax  —  by  —  c  is  an  equation  gf  the  first  degree  ; 
x2  -f  4  x  =  —  2,  an  equation  of  the  second  degree  ; 
2  x2  —  3  xy2  =  5,  an  equation  of  the  third  degree ;  etc. 
A   Linear,  or  Simple,  Equation   is  an  equation  of  the  first 
degree. 

26.  The  equations  of  Exercise  1  were  integral,  first  degree 
in  one  unknown  number,  linear. 

THEOREMS   IN   REGARD   TO   EQUIVALENT   EQUATIONS 

27.  If  the  same  expression  be  added  to  both  members 
of  an  equation,  the  resulting  equation  will  be  equivalent 
to  the  first.  ; 

Let  A=B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

To  prove  the  equation     A  +  C  =  B  +  C,  (2) 

where  C  is  any  expression,  equivalent  to  (1).  - 

Any  solution  of  (1),  when  substituted  for  the  unknown  numbers, 
makes  A  identically  equal  to  B  (§  15). 

It  then  makes  A  +  C  identically  equal  to  B  +  C  (§  16,  1). 

Then  it  is  a  solution  of  (2). 


12  ALGEBRA 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown  num- 
bers, makes  A+C  identically  equal  to  B  -f  C. 
It  then  makes  A  identically  equal  to  B  (§  16,  2). 
Then  it  is  a  solution  of  (1). 
Therefore,  (1)  and  (2)  are  equivalent. 
The  principle  of  §  16,  1,  is  a  special  case  of  the  above. 

28.  The  demonstration  of  §  27  also  proves  that 

If  the  same  expression  be  subtracted  from  both  mem- 
bers of  an  equation,  the  resulting  equation  will  be  equiva- 
lent to  the  first. 

The  principle  of  §  16,  2,  is  a  special  case  of  this. 

29.  If  the  members  of  an  equation  be  multiplied  by 
the  same  expression,  which  is  not  zero,  and  does  not 
involve  the  unknown  numbers,  the  resulting  equation 
will  be  equivalent  to  the  first. 

Let  A  =  B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

To  prove  the  equation      A  x  <7  =  B  x  (7,  (2) 

where  C  is  not  zero,  and  does  not  involve  the  unknown  numbers,  equiva- 
lent to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  numbers, 
makes  A  identically  equal  to  B. 

It  then  makes  A  x  C  identically  equal  to  B  x  C  (§  16,  3). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown  num- 
bers, makes  A  x  C  identically  equal  to  B  x  C. 

It  then  makes  A  identically  equal  to  B  (§  16,  4). 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  reason  why  the  above  does  not  hold  for  the  multiplier  zero  is, 
that  the  principle  of  §  10,  4,  does  not  hold  when  the  divisor  is  zero. 

The  principle  of  §  10,  3,  is  a  special  case  of  the  above. 

30.  If  the  members  of  an  equation  be  multiplied  by  an  ex- 
pression which  involves  the  unknown  numbers,  the  resulting 

equation  is,  in  general,  not  equivalent  to  the  first. 

Consider,  for  example,  the  equation  x  +  2  =  3  x  —  4.  (1) 

Now  the  equation 

(x  +  2) (x  -  1)  =  (3  x  -  4)<>  -  1),  (2) 


EQUIVALENT   EQUATIONS  13 

which  is  obtained  from  (1)  by  multiplying  both  members  by  x —  1,  is 
satisfied  by  the  value  x  =  1,  which  does  not  satisfy  (1). 
Then  (1)  and  (2)  are  not  equivalent. 

Titus  it  is  never  allowable  to  multiply  bath  members  of  an 
integral  equation  by  an  expression  which  involves  the  unknown 
numbers;  for  in  this  way  additional  solutions  are  introduced. 

31.  If  the  members  of  an  equation  be  divided  by  the 
same  expression,  which  is  not  zero,  and  does  not  involve 
the  unknown  numbers,  the  resulting  equation  will  be 
equivalent  to  the  first. 

Let  A=B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

A       B 

To  prove  the  equation  —  =  — ,  (2) 

0        0 

where  C  is  not  zero,  and  does  not  involve  the  unknown  numbers,  equiva- 
lent to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  numbers, 
makes  A  identically  equal  to  B. 

A  B 

It  then  makes  —  identically  equal  to  —  (§  16,  4). 

0  C 

Then  it  is  a  solution  of  (2). 
Again,  any  solution  of  (2),  when  substituted  for  the  unknown  num- 

A  B 

bers,  makes  —  identically  equal  to  —  • 

It  then  makes  A  identically  equal  to  B. 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  principle  of  §  16,  4,  is  a  special  case  of  the  above. 

32.  If  the  members  of  an  equation  be  divided  by  an  expres- 
sion which  involves  the  unknown  numbers,  the  resulting  equa- 
tion is,  in  general,  not  equivalent  to  the  first. 

Consider,  for  example,  the  equation 

O  +  2)(x  -  l)  =  (3x  -  4) (a  -  1).  (1) 

Also  the  equation  x  -}-  2  =:  Sx  —  4,  -^  (2) 

which  is  obtained  from  (1)  by  dividing  both  members  by  x  —  1. 

Now  equation  (1)  is  satisfied  by  the  value  x  =  1,  which  does  not  sat- 
isfy (2). 

Then  (1)  and  (2)  are  not  equivalent. 


I 


14  ALGEBRA 

It  follows  from  this  that  it  is  never  allowable  to  divide  both 
members  of  an  in  (earn  I  equation  by  an  expression  which  in- 
volves the  unknown  numbers  ;  for  An  this  way  solutions  are  lost. 

33.  If  both  members  of  a  fractional  equation  be  multi- 
plied by  the  L.C.M.  of  the  given  denominators,  the  re- 
sulting equation  is  in  general  equivalent  to  the  first. 

Let  all  the  terms  be  transposed  to  the  first  member,  and  let 
them  be  added,  using  for  a  common  denominator  the  L.  C.  M. 
of  the  given  denominators. 

The  equation  will  then  be  In  the  form 

-  =  0.  (1) 

We  will  now  prove  the  equation 

-4  =  0,  (2) 

which  is  obtained  by  multiplying  (1)  by  the  L.  C.  M.  of  the  given  denomi- 
nators, equivalent  to  (1),  if  A  and  B  have  no  common  factor. 

Any  solution  of  (1),  when  substituted   for  the  unknown   numbers, 

A 

makes  —  identically  equal  to  0. 

Then,  it  must  make  A  identically  equal  to  0. 

Then,  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  0. 

Since  A  and  B  have  no  common  factor,  B  cannot  be  0  when  this  solu- 
tion is  substituted  for  the  unknown  numbers. 

Then,  any  solution  of  (2),  when  substituted  for  the  unknown  numbers, 

A 

makes  —  identically  equal  to  0,  and  is  a  solution  of  (1). 
B 

Therefore,  (1)  and  (2)  are  equivalent,  if  A  and  B  have  no  common 
factor. 

If  A  and  B  have  a  common  factor,  (1)  and  (2)  are  not  equivalmt ; 
consider,  for  example,  the  equations 

£=JL  -=  0,  and  x  -  1  =  0. 
a*  - 1 

The  second  equation  is  satisfied  by  the  value  x  =  1,  which  does  not 
satisfy  the  first  equation  ;  then,  the  equations  are  not  equivalent. 


EQUIVALENT   EQUATIONS  15 

34.  A  fractional  equation  may  be  cleared  of  fractions  by 
multiplying  both   members  by  any  common   multiple   of  the 

denominators;  but  in  this  way  additional  solutions  are  often 
introduced,  and  the  resulting  equation  is  not  equivalent  to  the 
first. 

Consider,  for  example,  the  equation 

%        I       %      2 

x2  —  1      x  —  1 

If  we  solve  by  multiplying  both  members  by  x2  —  1,  the  L.  C.  M.  of 
x2  —  1  and  x  —  1,  we  find  x  —  —  2. 

If,  however,  we  multiply  both  members  by  (x2  —  l)(x  —  1),  we  have 

xs  —  x2  +  x3  —  x  —  2  x3  —  2  x2  —  2  x  +  2,  or  x2  +  x  —  2  =  0. 

The  latter  equation  may  be  solved  by  using  factors. 
The  factors  of  x2  +  x  —  2  are  x  +  2  and  x—1. 
Solving  the  equation  x  -f  2  =  0,  x  =  —  2. 
Solving  the  equation  x—  1=0,  x  =  1. 

This  gives  the  additional  value  x  =  1 ;  and  it  is  evident  that  this  does 
not  satisfy  the  given  equation. 

35.  If  both  members  of  an  equation  be  raised  to  the 
same  positive  integral  power  (§  66),  the  resulting  equa- 
tion will  have  all  the  solutions  of  the  given  equation, 
and,  in  general,  additional  ones. 

Consider,  for  example,  the  equation  x  =  3. 
Squaring  both  members,  we  have 

x2  =  9,  or  x2  -  9  =  0,  or  (x  +  3)  (x  -  3)  =  0. 

The  latter  equation  has  the  root  3,  and,  in  addition,  the  root  —  3. 
We  will  now  consider  the  general  case. 

Let  A  =  B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

Raising  both  members  to  the  »th  power,  n  being  a  positive  integer,  we 

have  An  =  £*,  or  A*  -  Bn  =  0.  (2) 

Factoring  the  first  number  (§  103,  VII), 

(A  -  B)  (An-*  +  A»~2B  +  •  •  •  +  5"-1)  =  0.  (3) 

Now,  equation  (3)  is  satisfied  when  A—B. 
Whence,  equation  (2)  has  all  the  solutions  of  (1). 
But  (3)  is  also  satisfied  when 


16  ALGEBRA 

An-l  +  An-TB  4.  ...  4.  7^-1  =  0  ; 

so  that  (2)  lias  also  the  solutions  of  this  last  equation,  which,  in  general, 

do  not  satisfy  (1). 

EQUIVALENT  SYSTEMS  OF  EQUATIONS 

36.  Two  systems  of  equations,  involving  two  or  more  un- 
known numbers,  are  said  to  be  equivalent  when  every  solution 
of  the  first  system  is  a  solution  of  the  second,  and  every  solu- 
tion of  the  second  is  a  solution  of  the  first. 

are  equations  involving  two  or  more  unknown  numbers, 
the  system  of  equations 

A  =  0, 
mA  +  nB  =  0, 

where  m  and  n  are  any  numbers,  and  n  not  equal  to 
zero,  is  equivalent  to  the  first  system. 

For  any  solution  of  the  first  system,  when  substituted  for  the  un- 
known numbers,  makes  A  =  0  and  B  —  0. 

It  then  makes  ^4  =  0  and  mA  +  nB  —  0. 

Then,  it  is  a  solution  of  the  second  system. 

Again,  any  solution  of  the  second  system,  when  substituted  for  the 
unknown  numbers,  makes  A  =  0  and  mA  +  nB  =  0. 

It  therefore  makes  nB  =  0,  or  B  =  Q. 

Since  it  makes  J.  =  0  and  B  =  0,  it  is  a  solution  of  the  first  system. 

Hence,  the  systems  are  equivalent. 

A  similar  result  holds  for  a  system  of  any  number  of  equations. 

Either  m  or  n  may  be  negative. 

38.  If  either  equation,  in  a  system  of  two,  be  solved 
for  one  of  the  unknown  numbers,  and  the  value  found  be 
substituted  for  this  unknown  number  in  the  other  equa- 
tion, the  resulting  system  will  be  equivalent  to  the  first. 

Let  ]— **  £ 

(2) 


f  A  =  B, 


be  equations  Involving  two  unknown  numbers,  t  and  f. 
Let  W  be  the  value  at  .<  obtained  by  solving  (1). 


EQUIVALENT  EQUATIONS  •    17 

Let  F  =  Q  be  the  equation  obtained  by  substituting  E  for  x  in  (2). 
To  prove  the  system  of  equations 

(x=E,  (3) 

1  F  =  G  (4) 

equivalent  to  the  first  system. 

Any  solution  of  the  first  system  satisfies  (3),  for  (3)  is  only  a  form  of  (1). 

Also,  the  values  of  x  and  y  which  form  the  solution  make  x  and  E 
equal ;  and  hence  satisfy  the  equation  obtained  by  putting  E  for  x  in  (2). 

Then,  any  solution  of  the  first  system  satisfies  (4). 

Again,  any  solution  of  the  second  system  satisfies  (1),  for  (1)  is  only 
a  form  of  (3). 

Also,  the  values  of  x  and  y  which  form  the  solution  make  x  and  E 
equal ;  and  hence  satisfy  the  equation  obtained  by  putting  x  for  E  in  (4). 

Then,  any  solution  of  the  second  system  satisfies  (2). 

Hence,  the  systems  are  equivalent. 

A  similar  result  holds  for  a  system  of  any  number  of  equations,  in- 
volving any  number  of  unknown  numbers. 

39.  The  principles  of  §§  27,  28,  29;  31,  33,  35,  36,  and  37 
hold  for  equations  of  any  degree. 

40.  In  the  solution  of  an  equation  of  Exercise  1,  we  replaced 
each  equation  by  an  equivalent  one  more  easily  solved  for  the 
unknown  number. 

41.  Elimination  is  the  process  of  deriving  from  a  system  of 
two  or  more  equations,  a  system  containing  one  less  unknown 
number  than  the  given  system. 

There  are  several  methods  of  elimination,  each  method  de- 
pending on  a  process  which  wrill  form  a  second  system  equiva- 
lent to  the  first. 

42.  A  system  of  equations  is  called  Simultaneous  when  each 
contains  two  or  more  unknown  numbers,  and  every  equation 
of  the  system  is  satisfied  by  the  same  set,  or  sets,  of  values 
of  the  unknown  numbers ;  thus,  each  equation  of  the  system 

Jx  +  y  =  6, 
ix  —  y  =  3, 

is  satisfied  by  the  set  of  values  x  =  4,  y  =  1. 


18  ALGEBRA 

A  Solution  of  a  system  of  simultaneous  equations  is  a  set  of 
values  of  the  unknown  numbers  which  satisfies  every  equation 
of  the  system ;  to  solve  a  system  of  simultaneous  equations  is 
to  find  its  solutions. 

Ex.     Solve  (1)     2x  +  6y  =  9,  ] 

(2)         x  -  y  =  1,  J 

(1)    2x  +  5y  =  %\ 

(8)     5x-5*/  =  5,  J 

-      0)  2*  +  6|f  =  9,        ]         2s  +  5y  =    9,1 

(4)     (2a;  +  5?/)+  5s-  by  =9  +  6,  j  '  7*  =  14,  J 

System  II  is  equivalent  to  system  I,  and  system  III  is 
equivalent  to  system  II. 

System  III  gives  the  required  solution  since  (4)  gives  x  =  2 
and  this  value  substituted  in  (1)  gives  y  =  1. 

Similarly  it  may  be  shown  that  elimination  by  substitution 
and  by  comparison  involve  the  deriving  of  equivalent  systems 
from  the  given  system  (§§  37,  38). 

Unless  the  equations  of  a  given  system  are  independent  a 
solution  is  not  possible. 

43.  If  two  equations,  containing  two  or  more  unknown  num- 
bers, are  not  equivalent,  they  are  called  Independent. 

Consider  the  equations 

(x  +  y  =  5,  (1) 

1  x  +  y  =  6.  (2) 

It  is  evidently  impossible  to  find  a  set  of  values  of  x  and  y  which  shall 
satisfy  both  (1)  and  (2). 

Such  equations  are  called  Inconsistent. 


EXERCISE  4 

Solve  the  following  equations,  using  Addition  or  Subtrac- 
tion, Substitution  or  Comparison : 

3x  +  5y  =  21.  2      <x-2y  =  9. 

x-2y=   8,  I  2  x-y  =  12. 


5- 
6. 

7- 
8. 


Ax-  3?/  =  1. 

6  a;  +  15?/  =  8. 

G  a?  +  9  =  3  ?/. 

y  =  4  +  x. 

3  a  -  3  #  =  -  12. 

3  x  —  -J-  y  =  17. 

3x-2y  =  l%. 

x=2y. 

i  ra  +  §  n  =  —  2. 

3  m  + 12  =  -  4  n. 

S  +  4:V=  —  1. 


EQUIVALENT   EQUATIONS 

+  2 


19 


1. 


10. 


ii. 


12. 


*3- 


=  ^s-16. 
f2 


M- 


15. 


16. 


i7- 


18. 


(x-3y)- 


2  x—  y 
2 


f-Sf 


[17  />  —  7 

1       '_     <=j>-3q. 

I  8j>  +  g«=15. 

[3^  +  "^zl/  =  25. 
3 

|l5-2a+^=0. 

5 


fll*=W  +  l& 

l2*-t*  =  10. 


5. 


2# 

+  3.v 

-6 

#  — 

•?/-l 

9 

7 

6  # 

—  5  y 

+  10 

5  a: 

+  3y 

7  15' 

'  y  —  3x  =  a. 
I  x  +  ^  1/  =  9  a. 

'a;  +  y  +  2      3x  —  y      ~  _x 
~~4  17  ~~6' 

9^5 

\  X  +  V  —  1        1/  N       i 


20  ALGEBRA 

19. 


2  a%  —  4  by  =  a2  —  ab  -f  2  62. 

gg-f-y. 


fr**-* 


-l=c  +  d.                                             I  a?       v       1Ao 
20.     J      4.  —  =  10f . 

[  2  #  —  y  =  5  c  —  Id. 

22.  If  5  in.  be  added  to  the  length  and  3  in.  to  the  breadth 
of  a  certain  rectangle,  the  area  is  increased  by  120  sq.  in.,  but 
if  1  in.  be  subtracted  from  the  length  and  2  in.  from  the  breadth, 
the  area  is  decreased  by  70  sq.  in.     Find  its  dimensions. 

23.  2  cu.  ft.  of  water  and  4  cu.  ft.  of  ice  together  weigh 
355  lb.  The  difference  between  the  weights  of  3  cu.  ft.  of 
water  and  2  cu.  ft.  of  ice  is  72  lb.  8  oz.  Find  the  weights  of  a 
cubic  foot  of  each. 

24.  A  masonry  contractor  held  back  $132.50  of  the  wages 
due  his  men.  His  bricklayers  earned  $  3  per  day,  and  his  hod 
carriers  $  1.75  per  day.  Their  combined  wages  for  a  day  were 
$  256.25.  He  retained  $  1.50  from  each  bricklayer  and  $  1  from 
each  hod  carrier.     How  many  carriers  did  he  employ  ? 

25.  A  man  rows  a  certain  distance  down  stream  at  the  rate 
of  33  mi.  an  hour  in  3^-  hr.  In  returning  it  takes  him  16  hr.  to 
reach  a  point  5  mi.  below  his  starting  point.  Find  the  rate  of 
the  current. 

26.  Two  trains  start  toward  each  other,  one  from  New  York, 
the  other  from  Chicago.  They  meet  in  10  hr.,  40  min.,  the 
distance  between  the  two  cities  being  960  mi.  If  the  first 
train  starts  3  hr.  earlier  than  the  second  train,  they  will  meet 
9£  hr.  after  the  second  train  starts.  Find  the  rate  of  each 
train. 

27.  A  number  lies  between  300  and  100.  If  18  is  added  to 
the  number,  the  last  two  digits  change  places  with  each  other, 
and  if  the  number  be  divided  by  the  number  expressed  by  the 
first  two  digits,  the  quotient  is  10^T.     Find  the  number. 


GRAPHICAL   REPRESENTATION  21 

28.  Find  two  numbers  whose  difference  is  93  and  whose  sum 
divided  by  the  smaller  number  gives  a  quotient  of  0:7\ 

29.  By  the  law  of  levers,  the  product  of  the  weight  Wx  by 
the  distance  from  Wx  to  the  fulcrum,  F,  is  equal  to  the  product 
of  the  weight  W2  by  the  dis-  ir,  Jr., 
tance  from  W2  to  the  fulcrum.                              ' 

A  board  resting  across  a  pole  balances  when  a  60-lb.  boy  is  on 
one  end  and  a  100-1  b.  boy  on  the  other  end.  The  board  will 
also  balance  if  a  120-lb.  boy  sits  2  ft.  from  one  end  and  a  60-lb. 
boy  sits  2  ft.  from  the  other  end.    Find  the  length  of  the  board. 

30.  If  a  regular  hexagon  is  circumscribed  about  a  given 
circle,  the  difference  between  the  areas  of  the  hexagon  and 
circle  is  32.24,  and  the  sum  of  their  areas  is  660.56.  Find  the 
radius  of  the  circle. 

GRAPHICAL,  REPRESENTATION 

44.  A  drawing  or  picture  of  given  data  or  of  an  equation  is 
often  of  value. 

45.  Descartes  (1596-1650)  was  the  first  mathematician  to 
apply  measurement  to  equations. 

It  is  impossible  to  locate  absolutely  a  point  in  a  plane.  All 
measurements  are  purely  relative,  and  all  positions  in  a  plane 
or  in  space  are  likewise  relative.  Since  a  plane  is  infinite  in 
length  and  breadth,  it  is  necessary  to  have  some  fixed  form 
from  which  one  can  take  measurements.  For  this  form, 
assumed  fixed  in  a  plane,  Descartes  chose  two  intersecting 
lines  as  a  coordinate  system.  Such  a  system  of  coordinates 
has  since  his  time  been  called  Cartesian.  It  will  best  suit  our 
purpose  to  choose  lines  intersecting  at  right  angles. 

46.  The  Point.  If  we  take  any  point  M,  its  position  is 
determined  by  the  length  of  the  lines  QM—x  and  PM?*p, 

parallel  to  the  intersecting  lines  OX  and  OF  (Fig.  £).  The 
values  x=  a  and  y  =  h  will  thus  determine  a  point.  The  unit 
of  length  can  be  arbitrarily  chosen,  but  when  once  fixed  remains 


22 


ALGEBRA 


Y 

Fig.  2. 


the  same  throughout  the  problem  under  discussion.     QM=  x 
and  PM=y,  we  call  the  coordinates  of  the  point  M.    x,  measured 

parallel  to  OX,  is  called  the  abscissa, 
y,  measured  parallel  to  OF,  is  the 
ordinate.  OX  and  OFare  the  coordi- 
nate axes.  OX  is  the  axis  of  x,  also 
called  the  axis  of  abscissas.  OF  is 
the  axis  of  y,  also  called  the  axis  of 
ordinates.  0,  the  point  of  intersec- 
tion, is  called  the  origin. 

Two  measurements  are  necessary 
to  locate  a  point  in  a  plane. 

For  example,  x  =  2  holds  for  any  point  on   the         ,     t     ty        a 
line  AB  (Fig.  2).     But  if  in  addition   we   demand 
that  y  =  3,  the  point  is  fully  determined  by  the  in- 
tersection of   the   lines   AB  and  CD,  any  point  on 
CD  satisfying  the  equation  y  =  3. 

47.   The     Line.       Consider     the     equation 

x  +  y  =  6. 

In  this  equation,  when  values  are  assigned  to  x,  we  get  a  value 
of  y  for  every  such  value  of  x.  When  x  =0,  y=6;  x=l,  ?/  =  5; 
x  —  2,  y  =  4  ;  x  =  3,  y=  3  ;  x  =  6,  y  =  1  ;  etc.,  giving  an  infinite  number 
of  values  of  x  and  2/  which  satisfy  the  equation. 

Laying  off  these  values  on  a  pair  of  axes,  as  shown  in 
§  46,  we  see  that  the  points  whose  coordinates  satisfy  this 
equation  lie  on  the  line  AB  (Fig.  3).  It  is  readily  seen  that 
there  might  be  confusion  as  to  the  direction  from  the  origin 
in  which  the  measurements  should  be  taken.  This  is  avoided 
by  a  simple  convention  in  signs.  Negative  values  of  x  are 
measured  to  the  left  of  the  ?/-axis,  positive  to  the  right.  In 
like  manner,  negative  values  of  y  are  measured  downward 
from  the  #-axis,  positive  values  upward.  XO  F,  VOX'.  X'OY', 
Y'OX,  are  spoken  of  as  the  first,  second,  third,  and  fourth 
quadrants  respectively.     (See  Fig.  2.) 

r>y  plotting  other  equations  of  the  first  degree  with  fcwo  un- 
known quantities  it  will  be  seen  that  such  an  equation  always 


GRAPHICAL    REPRESENTATION 


23 


represents  a  straight  line.  This  line  All  |  Fig,  3)  is  called  the 
graph  of  x  +  y=*6  and  is  the  locus  of  all  the  points  satisfying 
that  equation. 

48.  Now  plot  two  simultaneous  equations  of  the  first  degree 
on  the  same  axes,  e.g.  x  +  y  =  6  and  2  x  —  3  y  =  —  3  (  Fig.  1  ). 
We  see  that  the  coordinates  of  the  point  of  intersection  have 
the  same  values  as  the  x  and  y  of  the  algebraic  solution  of  the 
equations. 

This  is  a  geometric  or  graphical  reason  why  there  is  but  one 
solution  to  a  pair  of  simultaneous  equations  of  the  first  degree 
with  two  unknown  numbers.  A  simple  algebraic  proof  will 
be  given  in  the  next  article.  Hereafter  an  equation  of  the 
first  degree  in  two  variables  will  be  called  a  linear  equation. 


A> 

Y 

X 

o 

lB 

A 

Y 

s 

\ 

L) 

s 

v* 

y 

\ 

y 

/* 

J 

? 

\ 

* 

\ 

/ 

\ 

x 

0 

•* 

o 

\ 

B 

Fig.  3.  Fig.  4. 

49.  Algebraic  Proof  of  the  Principle  of  §48.  Two  simul- 
taneous equations  of  the  first  degree  cannot  be  satisfied  by  two 
different  sets  of  values  for  x  and  y.     Given  the  equations 

"•<•+  %  =?$,  (0 

ex+ft  =  h.  (2) 

Eliminating  ?/,                 («/—  ^V  =  <"/  —  hJl-  (%) 

Let  Xi  and  x2  be  the  roots  of  (3),  different  in  value.     Substituting 

these  roots,  we  have  (of—  eh)xi  =  cf  —  bh. 

(of-  eb)x2  =  cf-  bh\ 


But  X\  ^  .r2,  .'.  af- 


;,  whicli  is  impossible. 


In.  general,  the  plotting  of  two  graphs  on  the  same  Efcses  will 

determine  all  the  real   solutions   of   the  two  equations,   the 


24 


ALGEBRA 


(1) 


(2) 


coordinates  of  each  point  of  intersection  of  the  graphs  being 
values  of  x  and  y  which  satisfy  both  equations. 

50.  It  is  well  to  introduce  the  subject  of  graphs  by  the  use 
of  concrete  problems  which  depend  on  two  conditions  and 
which  can  be  solved  without  mention  of  the  word  equation. 

Professor  F.  E.  Nipher,  Washington  University,  St.  Louis,  proposes 
the  following : 

t;A  person  wishing  a  number  of  copies  of  a  letter  made,  went  to  a 
typewriter  and  learned  that  the  cost  would  be,  for  mimeograph  work  : 

1 1.00  for  100  copies, 
$2.00  for  200  copies, 
$3.00  for  300  copies, 
$4.00  for  400  copies,  and  so  on. 

"  He  then  went  to  a  printer  and  was  made  the  following  terms  : 
$2.50  for  100  copies, 
$3.00  for  200  copies, 
$3.50  for  300  copies, 

$4.00  for  400  copies,  and  so  on,  a  rise  of  50  cents 
for  each  hundred. 

"  Plotting  the  data  of  (1)  and  (2)  on  the  same 
axes,  we  have : 

M  The  vertical  axis  being  chosen  for  the  price-units, 
the  horizontal  axis  for  the  number  of  copies. 

"Any  point  online  (1)  will  determine  the  price 
for  a  certain  number  of  mimeograph  copies.  Any 
point  on  line  (2)  determines  the  price  and  cor- 
responding number  of  copies  of  printer's  work."' 

Numerous  lessons  can  be  drawn  from  this  problem.  One  is 
that  for  less  than  400  copies,  it  is  less  expensive  to  patronize 
the  mimeographer.  For  400  copies,  it  does  not  matter  which 
party  is  patronized.  For  no  copies  from  the  mimeographer, 
one  pays  nothing.  How  about  the  cost  of  no  copies  from  the 
printer?     Why? 

The  graph  offers  an  excellent  method  for  the  solution  of 
indeterminate  equations  in  positive  integers. 

Ex.     Solve   3x  +  4ysm22   for  positive   integers.      Plotting 

the  equation,  we  have 


INEQUALITIES 


25 


Y 

JL 

4 

1 

*.   * 

O 

Fig.  6. 

We  see  that  the  line  crosses  the  corner  of  a  square  only  when  x  —  2 
and  x  —  6.  For  all  other  integral  values  of  x,  y  is  fractional.  The  only 
positive  integral  solutions  are,  therefore,  x  =  2,  y  =  ±\  x  =  6,  |f=1- 
This  corresponds  to  the  algebraic  result. 


51-    In  the  equation  y  = — ,  y  is  dependent  on  x  for  its 

value.  That  is,  every  change  in  x  produces  a  change  in  y. 
When  two  quantities  are  so  related,  the  first  is  said  to  be  a 
Function  of  the  second.  Similarly  y  =f(x)9  read  y  is  a  function 
of  x,  means  that  y  is  equal  to  some  expression  in  x.  In  place 
of  the  equation  represented  by  Fig.  6  one  might  have' 


/(*)  = 


22-3x 


f(x)=8-2x, 
.F(a?)=±4-f-** 

Make  a  graph  of  each  of  these  two  functions  and  find  their 
point  of  intersection. 

EXERCISE   5 

i.  f(x)  =  lx-2±,  find  /(0),  /(I),  /(2),  /(-  4),  /(3f ). 

2.    4>  (*)  =  x2  -  2  x  + 1,  find  cf>  (0),  <£  (1),  <f>  (2). 

\ 

*The  /(x)  and  F(x)  mean  simply  different  functions  of  fcm  In  these 
same  equations  /(0)  means  the  value  of  the  function  when  0  is  substi- 
tuted for  x  in  /(&)  =  8  -  2  x,  namely,  /(0)  =  8.  Similarly /(l)  =  8-2(1) 
=  6, 


26 


ALGEBRA 


Solve  the  following  by  means  of  graphs: 
I  2  x  —  5  //  =  —  16. 
I  3  x  •+•  7  7/  =  5.  ( 

a?-5     2x-7/-l_2?/-2 


hi 


3 


*(y) 


+»  —  l 


4 


3  y  -02 


-44 


F(»).w 


LB 

_7y-24' 


10 

5a 

-19 

2- 

3 
-7a 

INEQUALITIES 

52.  The  Signs  of  Inequality,  >  and  <,  are  read  "  is  greater 
than  "  and  "  is  less  than"  respectively. 

Thus,  a  >  b  is  read  "  a  is  greater  than  b  " ;  a  <  b  is  read  "  a 
is  less  than  6." 

53.  One  number  is  said  to  be  greater  than  another  when 
the  remainder  obtained  by  subtracting  the  second  from  the 
first  is  a  positive  number. 

One  number  is  said  to  be  less  than  another  when  the  remain- 
der obtained  by  subtracting  the  second  from  the  first  is  a  neg<*~ 
tive  number. 

Thus,  if  a  —  b  is  a  positive  number,  a  >  b ;  and  if  a  —  h  is  a 
negative  number,  a  <  b. 

54.  An  Inequality  is  a  statement  that  one  of  two  expressions 
is  greater  or  less  than  another. 

The  First  Member  of  an  inequality  is  the  expression  to  the 
left  of  the  sign  of  inequality,  and  the  Second  Member  is  the 
expression  to  the  right  of  that  sign. 

Any  term  of  either  member  of  an  inequality  is  called  a  term 
of  the  inequality. 

55*  Two  or  more  inequalities  are  said  to  subsist  in  the  same 
sense  when  the  first  member  is  the  greater  or  the  less  in  both. 
Thus,  a  >  b  and  c  >  d  subsist  in  the  same  sense. 


INEQUALITIES  27 

PROPERTIES   OF   INEQUALITIES 

56.  An  inequality  will  continue  in  the  same  sense 
after  the  same  number  has  been  added  to,  or  subtracted 
from,  both  members. 

For  consider  the  inequality  a>b. 
By  §  53,  a  —  b  is  a  positive  number. 
Hence,  each  of  the  numbers 

{a  +  c)  — (&+.c),  and  (a  —  c)  —  (b  —  c) 
is  positive,  since  each  is  equal  to  a  —  b. 

Therefore,  a  +  c  >  b  -f  c,  and  a  —  c  >  b  —  c.  (§  53) 

57.  It  follows  from  §  56  that  a  term  may  be  transposed 
from  one  member  of  an  inequality  to  the  other  by  chang- 
ing1 its  sign. 

If  the  same  term  appears  in  both  members  of  an  inequality,  affected 
with  the  same  sign,  it  may  be  removed. 

58.  If  the  signs  of  all  the  terms  of  an  inequality  be 
changed,  the  sign  of  inequality  must  be  reversed. 

For  consider  the  inequality   a  —  b  >  c  —  d. 

Transposing  every  term,         d—  c>b  —  a.  (§57) 

That  is,  b  —  a  <  d  —  c. 

59.  An  inequality  will  continue  in  the  same  sense 
after  both  nembers  have  been  multiplied  or  divided  by 
the  same  positive  number. 

For  consider  the  inequality  a  >  b. 

By  §  53,  a  —  b  is  a  positive  number. 

Hence,  if  m  is  a  positive  number,  each  of  the  numbers 

mCa  —  b)  and  a  ~~  \  or  ma—  mb  and  — ,  is  positive. 

m  m      m 

Therefore,  ma>m?>,  and  —  >  — « 

m      m 

60.  It  follows  from  §§  58  and  59  that  if  both  members  of 
an  inequality  be  multiplied  or  divided  by  the  same  nega- 
tive number,  the  sign, of  inequality  must  be  reversed- 

61.  If  any  number  of  inequalities,  subsisting  in  the 
same  sense,  be  added  member  to  member,  the  resulting 
inequality  will  also  subsist  in  the  same  sense. 


28  ALGEBRA 

For  consider  the  inequalities  a  >  ft,  a'  >  ft',  a"  >  ft",  •••. 
Each  of  the  numbers,  a  —  ft,  a'  —  ft',  a"  —  ft",  •••,  is  positive. 
Then,  their  sum  a  —  ft  +  a'  —  ft'  +  a"  —  ft"  +  •••, 

or  a  +  a'  +  a"  +  •••  -  (ft  +  ft'  +  ft"  +  •••), 

is  a  positive  number. 

Whence,  a  +  a'  +  a"  +  •••  >  ft  +  ft'  +  b"  +  .... 

If  two  inequalities,  subsisting  in  the  same  sense,  be  subtracted  mem- 
ber from  member,  the  resulting  inequality  does  not  necessarily  subsist  in 
the  same  sense. 

Thus,  if  a  >  ft  and  a1  >  ft',  the  numbers  a  —  ft  and  a'  —  ft'  are  positive. 

But  (a  —  ft)  — (a'  —  ft'),  or  its  equal,  (a  —  a')  — (ft  -  ft'),  may  be  posi- 
tive, negative,  or  zero ;  and  hence  a—  a'  may  be  greater  than,  less  than, 
or  equal  to  ft  —  6'. 

62.  If  a  >  b  and  a'  >  ft',  and  each  of  the  numbers  a,  ar,  &,  b\ 
is  positive,  then  aa<  <bb'. 

Since  a'  >  ft',  and  a  is  positive, 

aa'>ab'  (§59).  (1) 

Again,  since  a>ft,  and  ft'  is  positive, 

aft'  >  ftft'.  (2) 

From  (1)  and  (2),  aa'  >  ftft'. 

63.  If  we  have  any  number  of  inequalities  subsisting  in  the 
same  sense,  as  a  >  b,  a'  >  ft',  a"  >  ft",  •••,  and  each  of  the  num- 
bers a,  a',  a",  •••,&,  6',  &",  •••,  is  positive,  then 

aa'a"  ...  >  bb'b"  .... 
For  by  §  62,.  aa'>bb'. 

Also,  a"  >  ft". 

Then  by  §  62,  aa'a"  >  ftft'ft". 

Continuing  the  process  with  the  remaining  inequalities,  we  obtain 
finally  aa'a"--  >  ftft'ft".-. 

64.  Examples. 

i.   Find  the  limit  of  x  in  the  inequality 


9x 
4x 

+  6y>lll. 
+  6y  =    ffi. 

6x 
6x 

5x> 

+  9y  = 

45,  a 

74. 

99. 

[^EQUALITIES  29 

Multiplying  both  members  by  3  (§  59),  we  have 

21x-23<2x  +  15. 
Transposing  (§  57),  and  uniting  terms, 
19x<88. 
Dividing  both  members  by  19  (§  59), 

x<2. 
( This  means  that,  for  any  value  of  x  <  2,  1  x  —  ^-*  <  —  -f  5.  ^ 

2.  Find  the  limits  of  x  and  y  in  the  following : 

3x  +  2p>37.  (1) 

2x-\-3y  =  33.  \2) 

Multiply  (1)  by  3, 
Multiply  (2)  by  2, 
Subtracting  (§  56), 

Multiply  (1)  by  2, 
Multiply  (2)  by  3, 
Subtracting,  —  5  y  >  —  25 

Divide  both  members  by  —  5,        y  <  5  (§  60) . 

(This  means  that  any  values  of  x  and  ?/  which  satisfy  (2),  also  satisfy 
(1),  provided  x  is  >  9,  and  y  <  5.) 

3.  Between  what  limiting  valnes  of  x  is  x2  —  4  #  <  21  ? 
Transposing  21,  we  have 

x2-4x<21,  if  x2-4x-21  <0. 

That  is,  if  (x  -f  3)  (x  —  7)  is  negative. 

Now  (x  -f  3)  (x  —  7)  is  negative  if  x  is  between  —  3  and  7  ;  for  if 
x  <  —  3,  both  x  -f  3  and  x  —  7  are  negative,  and  their  product  positive  ; 
and  if  x  >  7,  both  x  +  3  and  x  —  7  are  positive. 

Hence,  x2  —  4  x  <  21,  if  x  >  —  3,  and  <  7. 

EXERCISE  6 
Find  the  limits  of  x  in  the  following : 

1.  (4*  +  5)2-4<(8a;  +  5)(2a;+3). 

2 .  (3  x  +  2)(x  +  3)-  4  x>  (3  x -2)(x-  3)  +  36. 

3.  (x  +  4)(5  a  -  2)  +  (2  *  -  3)2>  (3  x  +  4)2-  78. 


30  ALGEBRA 

4.  (x-3)(x  +  4)(x-5)<(x  +  l)(x-2)(x-3). 

5.  a\x  - 1)  <  2  62(2  x  -  1)  -  a&,  if  a  -  2  &  is  positive. 

Find  the  limits  of  x  and  y  in  the  following : 

ox-\-  6  y  <  45.  [  7  *  —  4  y  >  11 . 

3  x  -  4  y  =  —  1 1 .  J  3 .  e  +  7  //  =  :  15 . 

8.  Find  the  limits  of  x  when 

3  a;- 11  <  24- 11  a,  and5s  +  23<20a;+& 

9.  If  6  times  a  certain  positive  integer,  plus  14,  is  greater 
than  13  times  the  integer,  minus  63,  and  17  times  the  integer, 
111  in  1  is  23,  is  greater  than  8  times  the  integer,  plus  31,  what  is 
the  integer? 

10.  If  7  times  the  number  of  houses  in  a  certain  village, 
plus  33,  is  less  than  12  times  the  number,  minus  82,  and  9 
times  the  number,  minus  43,  is  less  than  5  times  the  number, 
plus  61,  how  many  houses  are  there? 

11.  A  farmer  has  a  number  of  cows  such  that  10  times 
their  number,  plus  3,  is  less  than  4  times  the  number,  plus 
79;  and  14  times  their  number,  minus  97,  is  greater  than  6 
times  the  number,  minus  5.     How  many  cows  has  he  ? 

12.  Between  what  limiting  values  of  x  is  x2  -f  3  x  <  4  ? 

13.  Between  what  limiting  values  of  x  is  x2  <  8  x  —  1 5  '.' 

14.  Between  what  limiting  values  of  x  is  3  x2  +  19  as  <  —  20  ? 

65.    If  a  and  b  are  unequal  numbers, 
a*+V>  2ab. 

For  (a  -  ?0'2  >°  5  6*i  a'2  ~  2  <fh  +  l)2  >°- 

Transposing  -  2  ab,  .  a2  +  V2>2  ab. 

1.    Prove  that,  if  a  does  not  equal  3, 

(a-r-2)(tt-2)  >0a-  L8, 


INEQUALITIES  31 

By  the  above  principle,  if  a  does  not  equal  3, 
a2  +  9>6a. 

Subtracting  13  from  both  members, 

a2  -  4  >  (3  a  -  13,  or  (a  +  2)  (a  -  2)  >  0  a  -  13. 

2.   Prove  that,  if  a  and  &  are  unequal  positive  numbers, 

as  +  63  >  a2b  +  62a. 
We  have,  a2  +  b2>2  ab,  or  a2  -  ab  +  &2  >  a&. 

Multiplying  both  members  by  the  positive  number  a  +  &, 
a3+  &3>a2&  +  &2a. 

EXERCISE  7 
i.    Prove  that  for  any  value  of  x,  except  §, 
3a;(3a;-10)>-25. 

2.  Prove  that  for  any  value  of  x,  except  $-, 

*  4  o;(aj  -  5)  >  8  x  -  49. 

3.  Prove  that  for  any  values  of   a  and  b,  if  4  a  does  not 
equal  3  6,        (4a+ 3  6)(4  a  -  3&)>66(4a  -  36). 

4.  Prove   that   for  any  values  of  x  and  y,  if  o  x  does  not 
equal  4  y,  '  5x(ox-6y)>2  y(5  x-S  y). 

Prove  that,  if  a  and  b  are  unequal  positive  numbers, 

5.  a8b  +  ab*  >  2  a2b2. 

6.  a3+a2b+ab2+bs>2ab(a  +  b). 


32  ALGEBRA 

III.   EXPONENTS 

66.  An  Exponent  is  a  number  written  at  the  right  of   and 
above  a  number. 

It   is   customary  to   speak  of   the  number  as  raised  to  the 
power  indicated  by  the  exponent. 

67.  The  laws  we  shall  develop  are  to  hold  for  any  exponent, 
whether  integral,  fractional,  positive,  negative,  or  zero. 

68.  The  number  raised  to  the  power  is  called  the  Base. 

69.  Meaning  of  a  Positive  Integral  Exponent. 

a8  =  a  •  a :•  •  a. 
a4  =  a  •  a  •  a  •  a. 

Similarly  if  m  is  a  positive  integer, 

am  =  a  •  a  •  a  •  •  •  •  torn  factors. 

The   following   results  have  been  proved   to  hold   for  any 
positive  integral  values  of  m  and  n : 

amxan  =  am+n  (F.  C.)  *  (1) 

(a-)"  =  amn  (F.  C.)  (2) 

70.  Meaning  of  a  Fractional  Exponent. 

Let  it  be  required  to  find  the  meaning  of  0s. 
If  (1),  §  69,  is  to  hold  for  all  values  of  m  and  n, 

5  5  5  5,5,5  . 

a5  x  a*  x  a*  =  a*  *  *  =  a5. 

Then,  the  f&trc!  power  of  a*  equals  a5.    • 
Hence,  a*  must  be  the  cube  root  of  a5,  or  a3  =  ^a5. 
We  will  now  consider  the  general  case.         p 
Let  it  be  required  to  find  the  meaning  of  aq,  where  j>  and  q 
are  any  positive  integers. 

*  F.  C.  refers  to  Wells's  First  Course  in  Algebra. 


EXPONENTS  33 

If  (1),  §  69,  is  to  hold  for  all  values  of  m  and  n, 

P  P  P  *  +  ?  +  ?+. "to*  term*  lXq 

aq  xaq  xaq  x  ••  •  to  q  factors  =  aq    9   q  =  aq     =  a*\ 

7' 

Then,  the  gth  power  of  aq  equals  oF. 

p  p 

Hence,  aq  must  be  the  qth.  root  of  ap,  or  aq  =  ^/^p9 

Hence,  in  a  fractional  exponent,  the  numerator  denotes 
a  r5ower,  and  the  denominator  a  root. 

For  example,  a*  =  -\/a?-,  b^  =  V&5;  x*  =-%/x\  etc. 

A  Surd  is  the  indicated  root  of  a  number,  or  expression, 
which  is  not  a  perfect  power  of  the  degree  denoted  by  the 
index  of  the  radical  sign ;  as  V2,  -\/59  or  v^  -}-  y.  , 

The  degree  of  a  surd  is  denoted  by  its  index ;  thus,  V«5  is  a 
surd  of  the  third  degree. 

A  quadratic  surd  is  a  surd  of  the  second  degree. 

71.  Meaning  of  a  Zero  Exponent. 

If  (1),  §  69,  is  to  hold  for  all  values  of  m  and  n,  we  have 

am  x  a0  =  am+0  =  am. 

Whence,  a0  =  —  =  1. 

am 

We  must  then  define  a0  as  being  equal  to  1. 

72.  Meaning  of  a  Negative  Exponent. 

Let  it  be  required  to  find  the  meaning  of  a~3. 
If  (1),  §  69,  is  to  hold  for  all  values  of  m  and  n, 

a~3  x  a3  =  a"3+3  =  a°  =  1  (§  71). 

Whence,  a-3  =  —  • 

a' 

We  will  now  consider  the  general  case. 

Let  it  be  required  to  find  the  meaning  of  a~%  where  s  repre- 
sents a  positive  integer  or  a  positive  fraction. 


34  ALGEBRA 

If  (1),  §  69,  is  to  hold  for  all  values  of  m  and  n, 
a~8  xa8  =  a~s+8=a()  =  1  (§  71). 

Whence,  a~8  =  —  • 

a8 

We  must  then  define  a~8  as  being  equal  to  1  divided  by  a8. 

For  example,  a~2  =  ~;  a~i  =  — ;  3x~1y~^  = — -;  etc., 
a"  a*  xy* 

73.    It  follows  from  §  72  that 

Any  factor  of  the  numerator  of  a  fraction  may  be 
transferred  to  the  denominator,  or  any  factor  of  the 
denominator  to  the  numerator,  if  the  sign  of  its  ex- 
ponent be  changed. 

Th  a2bs  =     bs     ^aWc-1  ^a2(T4     te 

US'  cd4      a~2cd4         d*      ~  b~3c ' 


EXERCISE   8 

Express  with  positive  exponents : 

i.    a~2bs.                        5.    3xyz~2.  9.  7  x4y~2z. 

2.  xiy~2z.'                     6.    5c~^dk  10.  4  a~Gb~8ck 

3.  2m~4w.                    7-    a-2xy~\  11.  8wV. 

4.  a~lb4c-\                   8.    3jr*?*.  12.  rV~*f  i 

Transfer   all  literal  factors   from  the  denominators   to  the 
numerators: 


6x* 

I3-  y 

16. 

1 

2  a26~3 

19. 

5  a*6~* 
9  c"3d^ 

mn~A 
I4'     3^' 

i7- 

a4 
cd"3  * 

20. 

4  in"W 

**   *£■ 

18. 

3  <?/-*. 

7ar^j 

xy2z 

4*4 

EXPONENTS  35 

Transfer   all   literal   factors    from    the    numerators   to   the 
denominators : 


21. 

asb2 
x*  ' 

24. 

Q       3 

5X3 

27- 

a~2 
b2c  ' 

22 

1  x2y~\ 

mPn* 
3  ab2c~l 

5d"4 

25- 
26. 

m6 

28. 

3  i)Jn~$ 

4  xy~lz2 

23- 

~  3 
n~'r2 

8  x~*y# 
3cd2 

74.    Proof  that  am  •  an  =  am+n  holds  for  all  values  of  m  and  n. 

p  v 

I.  Let  m  —  -  and  ??  =  -,  where  p,  </,  r,  and  s  are  positive 

integers.  -* 

We  have,  a*xa«  =  a*  X  a*  =  %/a»°  X  Va«r  (§  70) 

ps+qr  p    r 

=  %/aps  x  a?r  ==  9i/ap8+qr  =  a  «•     (§  70)  =  a«  •. 

We  have  now  proved  that  (1),  §  69,  holds  when  m  and  n  are 
any  positive  integers  or  positive  fractions. 

II.  Let  m  be  a  positive  integer  or  fraction ;  and  let  n  =  —  q, 
where  q  is  a  positive  integer  or  fraction  less  than  m. 

By  §  74,  I,  am~q  x  aq  =  am~q+q  =  am. 

Whence,  am~q  =  ^-  =  amxa~q  (§  73). 

aq  J 

That  is,  am  x  a~q  =  am~q. 

III.  Let  m  be  a  positive  integer  or  fraction  ;  and  let  n  =  q, 
where  q  is  a  positive  integer  or  fraction  greater  than  m. 

By  §  73,  am  x  a~q  =  -i-  =  -i-  (§  74,  II)  =  am~*. 

IV.  Let  m—  —p  and  n=  —  q,  where  p  and  q  are  positive 
integers  or  fractions. 

Then,  a-pXcrq  =  ^—  =  —  (  74,  I)  =  a"^. 

Then,  am  x  an  =  am+n  for  all  positive  or  negative,  integral  or 
fractional,  values  of  m  and  n. 


36  ALGEBRA 

EXERCISE  9 

Multiply  the  following : 
i.    a8  by  <H.  4-   2</W*  by  </tfb. 

3  \ 

2.  3  xly'h  by  aT^z3.  5-   ^2/  by  -37—  p 

#  %  2 

3.  2  c*d  by  ST-\/c5*  6.   a46c*  by  afc-V*. 

7.  3  x~y  by  -  2  afy;!*. 

8.  x*  +  »^  +  y^  by  z*  —  jff. 
9<   3  x  _  1  +  ar1  by  5  a  +  2. 

10.  «*— 2^'2/r+2/2  by  V#-f  V#- 

11.  x2  +  2  x  —  ar1  +  1  by  x  +  ar1  +  1. 

12. p*  +  x  —  V#  +  1  by  -y/x  +  1- 

a:"2 

Divide  the  following : 

13.  a2  by  a5.  l6  8  aib~i  $j  by  2  a-2fo3. 

14.  ar*?/  by  aj~3?/2.  17.  5  mAn~%  by  2  ra~%*. 

15.  3y/xy3  by  x-*y-\  18.  a+2a*6*  +  &  by  a?+$. 

19.  m~*  +  3  mrhfc  +  3  m"^i  +  »'  by  m""2  +  ti*. 

20.  6  x2  -  6  x-2  -  12-  11  a;  +  23  ar1  by  2  x  +  1  -  3  x~\ 
2i.9  a;*?/-1  -  6  x%y-2  -5  +  12  x'hj  +  4  afty  by  3  a;  +  4^aTi/2 
22.   2^-7  a*  +  4  af*  -  11  +  12  a?*  by  2  a?*  +  4  -  3  xK 
Find  the  values  of  the  following: 

23-  ^)_\  *    (8*)1-  ,  29.  (.*fcfS 

24.  (m--)-'.  27.    (-27)*. 

1 

25.  (-v'aPp.  28.    (8~*)4,  30.    (VV)    v- 


EXPONENTS  37 

75.    To  prove  the  result 

(ab)n  =  anbn, 

for  any  fractional  or  negative  value  of  n. 

The  proof  of  this  result  in  the  case  where  n  is  any  positive 
integer,  was  given  in  F.  C. 

I.  Let  n  =s  where  p  and  q  are  any  positive  integers. 

[(o&)i]«  =  (aby  =  a?b>  (F.  C).  (1) 

p  p  p       p 

(a~«b~«)q  =  (a*)q(b*y  =  ap&*.  (2) 

From  (1)  and  (2),  [(a&)*]«  =  (a*6«)«. 

Taking  the  gth  root  of  both  members,  we  have 

p        p  p 
(ab)q  =  a*b«. 

II.  Let  n  =  —  s,  where  s  is  any  positive  integer  or  positive 
fraction. 

Then,  (a&)-=  -i-  =  -L(t  M,  JV)  =  o-*^. 

(ab)s      a8bs 

EXERCISE   10 
Find  the  values  of  the  following : 

1.  (asb^)4.  6.    (25a4)~*. 

1    \_f  7.    (32al\/F^)i 

2.  ' 


x'Zy    J  8.    (343  Vc-'SyK 

3.  (p'V'T1.  f%xf^ 

\  16  m~4 

4.  (a-V^c-1)-", 

10. 


5.    (Vafy-*)*.  \4afV5 


38  ALGEBRA 

EXERCISE   11 

Illustrative  Examples. 

Ex.  1.    Reduce  V|  to  its  simplest  form. 

A  surd  is  said  to  be  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  rational  and  integral,  is  not  a  perfect 
power  of  the  degree  denoted  by  any  factor  of  the  index  of  the 
surd,  and  has  no  factor  which  is  a  perfect  power  of  the  same 
degree  as  the  surd. 

§  =  28.  To  be  a  perfect  square  the  exponents  of  the  factors  of  the 
denominator  must  be  even  numbers.  Hence  multiplying  both  terms  of 
the  fraction  by  2,  we  have, 

vTI 

T' 


Vs=Vtss 


Ex.  2.   Reduce  V25  to  its  simplest  form. 


■v/25=\/V25=  V5. 

Ex.  3.   Express  5  V7  entirely  under  the  radical  sign. 

5V7=  V5V7,  or  (52)  1(7)1- 
By  §  75,      (52)i(7)l  =  (52 .  7)1  =  V175. 

Ex.  4.   Reduce  (5)1,  -\/3  to  the  same  degree. 

The  L.  C.  M.  of  the  indices  of  the  roots  is  12.     Hence, 

r^  =  ^/  3*  =  3i\ 

The  surds  are  now  of  the  twelfth  degree. 

Ex.  5.   Find  the  product  of  V45  and  V72. 
(45)1  (72)1  =  (32 .  6)1 .  (3j  .  23)i  _  (22 .  34 . 5  . 2)* 

=  2.3'2(5.2)2  =  18Vl0. 
Reduce  the  following  to  their  simplest  form  : 
*■    -v^-  4-    (72)1.  7.    5(32  aV//  >'. 

2.  (27  a8)*.  5.    ^128  a4ft*.  8.    7(80)1. 

3.  (45)1.  6.    3V25()a2.f.  9.    (CSC,)'. 


EXPONENTS  39 

10.  4\Vl86\  I3.  (x-yXa&m8)*. 

ii.  V4a2-5arty.  14.  (4  x2  -  24  xb  +  36  brf. 

12.  (a3-2a26-fa62)l  15.  V(a?+3a?+2)(a?+6a?+8). 

**  (*)*•                      18.  (i)*.                      20.   2V^. 

17.  (♦)*.  19.  Ki)*-  2I   /8a«\4 


75; 
22.  1(1^)K  24.  J^ZI. 

mV25n2y  *p-2q 

,     t         \(    1    \4  **  !       A«2-&2\} 

Express  entirely  under  the  radical  sign : 

26.  2V5.  ,    '  ■     /     x '  \i 

31.     (a?  +  y) )    . 

27.  3(2)*.  V-y-J 

28.  a(bc2y.  32.    Vm2  -f  mn  —  2  n\ 

_____  m  —  n 

29.  5  x  V3  #y.  , 

c  +  4  /c2  +  5c-   6V 

30.  (a  +  8ft)f— 1— V-  33'    c-1^4-8c  +  16y  ' 

\a  +  o  bj 

Reduce  the  following  to  equivalent  surds  of  the  same  degree : 
34.    V3,  a/4.  37.   5Vx,  3  Vary. 

35.  (2)4,  (i)i  (5)i.  3»-  («-»)*,  («+*)*;  &+*fy. 

36.  .(afy)*,  («y)*,  (®Y)*.  39-    v^  +  y8,  Vz4-#4. 

Simplify  the  following : 

40.  (18)4  +  3(50)4-2(72)4.   42.    V^  +  V'250  -2\Vl28. 

41.  2(27)4-5(48)4+11(75)4.   43.   f  (12)*- f(*)*  +  (3)i 

44.    8^80-2^/405  +  18^. 


40  ALGEBRA 

45.    (24  a*x)  *  +  2  (54  ax) '  -  5  (6  a*x)  • . 
fatmFY     famn*V  .  /aV3\* 

46'  [*r]  "fawj  + W  ' 

47-    a)i  +  3(TV)i  +  ^V56. 

48.    ^^^~3(4a2x2)*  +  5-v/8^V. 

(a  +  2/)2  x2-y\      x      J 

Multiply  the  following: 

50.  V90  by  V63.  56.    Va3^  by  ^ax^f. 

51.  (35)*  by  (105)*.  ,  57.    2(5)*  by  3(15)1 

52.  -^54  by  -v^S.  58.    5^40  by  6(o)\ 

53.  (3)*  by  (2)1  ftayi    ,jtf\l    /2y\1 

54.  a/2  by -^4.  W    A27&;     \15aJ  ' 
,55-  (*)*  by  (I)*  by  (f)*.  60.    VS^S  •  (2  a2)*  .  (6  r5)*. 

61.  3 (2)*  -  5  (3)4  by  4  (2)*  +  3  (3)1 

62.  5  V7  +  6  V2  by  V7  -  4  V2. 

63.  2  (8  a;)*  -  9  (2  #)*'  by  (2  a) *  -  3 (2  y)*. 

64.  3 (a  - 1)*  +  4(2  a  +  5)*  by  2 (a-  1)*  -  10(2  a  +  5)*. 

65.  5Vf-2V|by4Vf  +  9V|. 

Divide  the  following : 

66.  V72byV6.  7o.    (8  a*)'*  by  (16  a4)*. 

67.  2Vl25by4V5.  7?-    V722a? by  V2j^ 

.  72.    (4)*  by  (*)*. 

68.  (192)*  by  (12)*.  •  ,—1    T/    ,,_ 
V       ;      y  ^     ;                        73.    8  VaV  by  6</ax*. 

69.  (512)*  by  (16)*.  74.    (H)*  by  (9)5. 


EXPONENTS  41 

Find  raises  of  the  following: 

75-    (?VS)*.  80.    V  f/a*  -  2  a/;+7?'. 

76.    [1(4)*]*.  8l.   [8f^(5)ij, 

77-    (3s/2~o^)4.  82.    (4V5-2V7/-'. 

78.  t(S)¥-  83.  VK>)l-*(?)l^(*r  +  :>(^l 

79.   [(3a)ij*  84.    (7Vn-5V3)(7Vn  +  5V:V). 

Express  each  of  the  following  with  a  rational  denominator : 

85.  4-       '  89.    !±2& 
V2  v     2-V3 

86.  1*.  9o.    3(4)4^2^. 
V5  2(3)*+ (10)* 

^>2+»*  9       (a  + 6)*- («_*)* 

88.  a  +  ^-  92.   (^  +  ^  +  (^2-.y2P 

a  -  6*  (a2  +  y*)*  -  (x*  -  y2)* 

LOGARITHMS 

76.  Any  number  may  be  expressed  as  a  power  of  some  num- 
ber chosen  as  base. 

For  example,  4  =  22,  8  =  23,  64  =  26,  etc.  Numbers  between 
4  and  8  would  be  expressed  by  2n  where  rt  is  2  plus  some  frac- 
tional number.  In  suclr  a  case  the  exponent  is  called  the 
Logarithm  of  the  Number  to  the  Base  2. 

E.g.  2  is  the  logarithm  of  4  to  the  base  2 ;  3  is  the  logarithm  of  8  to 
the  base  2,  etc. 

77.  The  Common  System  of  logarithms  has  10  for  its  base. 
Every    positive    arithmetical    number    may   be    expressed, 

exactly  or  approximately,  as  a  power  of  10. 

Thus,  100  =  102;  13  =  1011139-;  etc. 


42  ALGEBRA 

When  thus  expressed,  the  corresponding  exponent  is  called 
its  Logarithm  to  the  Base  10. 

Thus,  2  is  the  logarithm  of  100  to  the  base  10 ;  a  relation 
which  is  written  log  10 100  =  2,  or  simply  log  100  =  2.     • 

Logarithms  of  numbers  to  the  base  10  are  called  Common 
Logarithms,  and,  collectively,  form  the  Common  System. 

They  are  the  only  ones  used  for  numerical  computations. 

78.  Any  positive  number,  except  unity,  may  be  taken  as  the 
base  of  a  system  of  logarithms ;  thus,  if  ax  =  m,  where  a  and 
m  are  positive  numbers,  then  x  =  logft  m. 

A  negative  number  is  nut  considered  as  having  a  logarithm. 

70.   By  §§  71  and  72, 

10°  =  1,  io-1^  —  =  .1, 

10 

10* =10,  io-«  =  A-  =  .oi, 

102  =  100,  10-3  =  -i- -  =  .001,  etc. 

103  ' 

Whence,  by  the  definition  of  §  76, 

log  1  =  0,  log  .1  =  - 1  =  9  -  10, 

log  10  =  1,  log  .01  =  -  2  =  8  -  10, 

log  100  =  2,  log  .001  =  -  3  =  7  - 10,  etc. 

The  second  form  for  log.l,  log. 01,  etc.,  is  preferable  in  practice. 
If  no  base  is  expressed,  the  base  10  is  understood. 

80.  It  is  evident  from  §  79  that  the  common  logarithm  of 
a  number  greater  than  1  is  positive,  and  the  logarithm  of  a 
number  between  0  and  1  negative. 

81.  If  a  number  is  not  an  exact  power  of  10,  its  common 
logarithm  can  only  be  expressed  approximately;  the  integral 
part  of  the  logarithm  is  called  the  characteristic,  and  the  deci- 
mal part  the  mantissa. 

For  example,  log  13  =  1.1139. 

Here;  the  characteristic  is  1,  and  the  mantissa  .1139. 


-EXPONENTS  43 

A  negative  logarithm  is  always  expressed  with  a  positive 
mantissa,  which  is  done  by  adding  and  subtracting  10. 

Thus,  the  negative  logarithm  —  2.5863  is  written  7.4137  —  10. 
In  this  case,  7  —  10  is  the  characteristic. 

The  negative  logarithm  7.4187  —  10  is  sometimes  written  3.4137 ;  the 
negative  sign  over  the  characteristic  showing  that  it  alone  is  negative,  the 
mantissa  being  always  positive. 

For  reasons  which  will  appear,  only  the  mantissa  of  the 
logarithm  is  given  in  a  table  of  logarithms  of  number ;  the 
characteristic  must  be  found  by  aid  of  the  rules  of  §§  82 
and  83. 

82.  It  is  evident  from  §  79  that  the  logarithm  of  a  number 
between         ±  and      10  is  equal  to  0  +  a  decimal ; 

10  and    100  is  equal  to  1  +  a  decimal ; 
100  and  1000  is  equal  to  2  -f-  a  decimal ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  number 
with  one  place  to  the  left  of  the  decimal  point  is  0 ;  with  two 
places  to  the  left  of  the  decimal  point  is  1 ;  with  three  places 
to  the  left  of  the  decimal  point  is  2 ;  etc. 

Hence,  the  characteristic  of  the  logarithm  of  a  number 
greater  than  1  is  1  less  than  the  number  of  places  to  the 
left  of  the  decimal  point. 

For  example,  the  characteristic  of  log  906328.51  is  5. 

83.  In  like  manner,  the  logarithm  of  a  number  between 

1  and      .1  is  equal  to  9  +  a  decimal  —  10 ; 
.1  and '  .01  is  equal  to  8  -f  a  decimal  —  10 ; 
.01  and  .001  is  equal  to  7  -j-  a  decimal  —  10 ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  decimal 
with  no  ciphers  between  its  decimal  point  and  first  significant 
figure  is  9,  with  —10  after  the  mantissa;  of  a  decimal  with 
one  cipher  between  its  point  and  first  significant  figure  is  8, 
with  —10  after  the  mantissa;  of  a  decimal  with  two  ciphers 
between  its  point  and  first  significant  figure  is  7,  with  —  10 
after  the  mantissa;  etc. 


44  ALGEBRA 

Hence,  to  find  the  characteristic  of  the  logarithm  of  a 
number  less  than  1,  subtract  the  number  of  ciphers  be- 
tween the  decimal  point  and  first  significant  figure  from 
9,  writing  —  10  after  the  mantissa. 

For  example,  the  characteristic  of  log  .007023  is  7,  with  —  10 
written  after  the  mantissa. 

PROPERTIES   OF   LOGARITHMS 

84.  In  ami  system,  the  logarithm  of  1  is  0. 

For  by  §  71 ,  cfi**  1  ;  whence,  by  §  78,  logal  =  0. 

85.  In  any  system  the  logarithm  of  the  base  is  1. 
For,  a1  =  a ;  whence,  loga  a  =  1. 

86.  In  any  system  whose  base  is  greater  than  1,  the  logarithm 

of  0  is  —  oo  .* 

For  if  a  is  greater  than  1,  a_Q0  =  —  =  —  —  0.     (The  discns- 

«*  .   oo 

sion  of  this  form  will  be  found  in  §  127.) 
Whence,  by  §  78,  log„  0  =  —  go. 

No  literal  meaning  can  be  attached  to  such  a  result  as  loga  0  =  —  oo  ;  it 
must  he  interpreted  as  follows  : 

If,  in  any  system  whose  base  is  greater  than  unity,  a  number  approaches 
the  limit  0,  its  logarithm  is  negative,  and  increases  indefinitely  in  abso- 
lute \alue. 

87.  In  any  system,  the  logarithm  of  a  product  is  equal  to  the 
sum  of  the  logarithms  of  its  factors. 

Assume  the  equations 

ax  =  m\  faj  =  logrtw, 

[;  whence,  by  §  <8,  { 
a*  =  n  ]  [//  =  logan. 

Multiplying  the  assumed  equations, 

ax  x  av  =  mn,  or  ax+y  =  mn. 
Whence,  loga  mn  =  x  -f-  y  =  loga  m  -+-  log,,  ». 

*  oo  stands  for  a  number  greater  than  any  aarigned  number.    Sec  §  l  *J<>. 


EXPONENTS  \~> 

In  like  manner,  the  theorem  may  be  proved  for  the  product 
of  three  or  more  factors. 

By  aid  of  §  87,  the  logarithm  of  a  composite  number  may 
be  found  when  the  logarithms  of  its  factors  are  known. 

Ex.    Given  log  2  =  .3010,  and  log  3  =  .4771 ;  find  log  72. 

log  72  =  log  (2  x  2  x  2  x  3  x  3) 

=  tog 2  -f  log2  4-  log2  4-  log3  4-  log3 

=  3  x  log 2  4-  2  x  log3  =  .9030  4-  .0542  =  1.8572. 

EXERCISE  12 

Given  log  2  =  .3010,  log  3  =  .4771,  log  5  =  .6990,  log  7  =  .8451, 
find : 

i.  log  15.        4.  log  125.  7.  log  567.  10. -log  1875. 

2.  log  98.         5.  log  315.  8.  log  1225.         11.  log  2646. 

3.  log  84.        6.  log  392.  9.  log  1372.         12.  log  24696. 

88.  In  any  system,  the  logarithm  of  a  fraction  is  equal 
to  the  logarithm  of  the  numerator  minus  the  logarithm 
of  the  denominator. 

Assume  the  equations 

ax  =  ml        ,  [#  =  logara, 

\  ;  whence,    \ 
ay  =  n  J  [y  =  \ogan. 

Dividing  the  assumed  equations, 

—  =  —    or  ax  y  =  —  • 
ay      n'  n 

Whence,  loga  -  =  x  —  y  =  logam  —  loga/i. 


Ex.    Given  log  2  =  .3010;  find  log  5. 

log  5  =  log  i?  =  log  10  -  log  2  =  1  -  .3010  =  .6990, 


46  ALGEBRA 

EXERCISE  13 

Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find  : 
i.    log-1/.  4-    log  245.  7.    log  |f.  10.    log-3™- 

2.  log-2f.  5.   log85f.  8.    log  375.  11.    log  46f . 

3.  log  11$.         6.   log  175.  9.    log  |f  12.   log  2ft- 

89.  In  any  system,  the  logarithm  of  any  power  of  a 
number  is  equal  to  the  logarithm  of  the  number  multi- 
plied by  the  exponent  of  the  power. 

Assume  the  equation  ax  =  m ;  whence,  x  =  loga  m. 
Raising  both  members  of  the  assumed  equation  to  the  jith 
power,        aPX  _  mP .  wjience?  i0ga  mv  —  pX  —  p  i0go  m 

90.  In  any  system,  the  logarithm  of  any  root  of  a  num- 
ber is  equal  to  the  logarithm  of  the  number  divided  by 
the  index  of  the  root. 

For,  logaVm  =  logtt(mr)  =  -  logam(§  89). 

91.  Examples. 

1.  Given  log  2  =  .3010 ;  find  log  2*. 

log  I*  =1*  log  2  =  -  x  .3010  =  .5017. 
o  o 

To  multiply  a  logarithm  by  a  fraction,  multiply  first  by  the  numerator, 
and  divide  the  result  by  the  denominator. 

2.  Given  log  3  =  .4771 ;  find  log  V3. 

log  ^  =  !°£»  =  i™  =.0506. 

8  8 

3.  Given  log  2  =  .3010,  log  3  =  .4771,  find  log  (2*  x  3*). 
By  §  87,  log  (2*  x  3*)  a  log  2*  +  log  3* 

=  -  log  2  +  -  log  3  =  .1003  +  .5964  =  .6067. 


EXPONENTS  47 

EXERCISE  14 

Given  log  2  =  .3010,  log  3  =  .4771,  Log  7  =  .g4&,  find; 


I. 

log  28. 

5- 

log  42". 

9- 

log 

oOl 

13-    log  {  s. 

2. 

log  57. 

6. 

log  45*. 

10. 

log 

(ft 

14.   log  V54. 

3- 

log  3*. 

7- 

log  63*. 

11. 

log  vs. 

15.   log  \  225. 

4- 

log  7 1 

8. 

log  98*. 

12. 

log 

^7. 

16.    log  v  1(32. 

i7- 

18. 

log  \/|. 

log(f)*. 

• 

21.   log 

^2' 

23- 

loew 

19. 
20. 

log  (3*  x  100*). 

log  (5a/3). 

22.    log 

2i 

o6 

24. 

log    £   . 

92.  In  the  common  system,  the  mantissae  of  the  log- 
arithms of  numbers  having  the  same  sequence  of  figures 
are  equal. 

Suppose,  for  example,  that  log  3.053  =  .4847. 

Then,  log  305.3  =  log(100  x  3.053)  =  log  100  +  log  3.053 
=  2  +.4847  =  2.4847  ; 
log  .03053  =  log  (.01  x  3.053)  =  log  .01  +  log  3.053 
=  8  -  10  +  .4847  a  8.4847  -  10  ;  etc. 

It  is  evident  from  the  above  that,  if  a  number  be  multiplied 
or  divided  by  any  integral  power  of  10,  producing  another 
number  with  the  same  sequence  of  figures,  the  mantissae  of 
their  logarithms  will  be  equal. 

For  this  reason,  only  mantissae  are  given,  in  a  table  of  Com- 
mon Logarithms;  for  to  find  the  logarithm  of  any  number,  we 
have  only  to  find  the  mantissae  corresponding  to  its  sequence 
of  figures,  and  then  prefix  the  characteristic  in  accordance 
with  the  rules  of  §§  82  and  83. 

This  property  of  logarithms  only  holds  for  the  •common 
system,  and  constitutes  its  superiority  over  other  systems  for 
numerical  computation. 


48  ALGEBRA    . 

93.  Ex.   Given  log  2  =.3010,  log  3  =.4771 ;  find  log  .00432. 

We  have  log  432  =  log  (2*  x  8*)  =  4  log  2  +  3  log  3  =  2.0353. 
Then,  by  §  92,  the  mantissa  of  the  result  is  .6858. 
Whence,  by  §  83,  log  .00432  =  7.0353-10. 

EXERCISE  15 
Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find  : 

1.  log  2.7.                  6.   log  .00000680.  n.  log  337.5. 

2.  log  14.7.                7.   log  .00125.  12.  log  3.888. 

3.  log  M.                 8.   log  5670.  13.  log  (4.5)8. 

4.  log  .0162.              9.   log  .0000588.  14.  log  -y/SA. 

5.  log  22.5.              10.    log  .000864.  15.  log  (24.3)1 

USE   OF   THE   TABLE 

94.  The  table  (pages  50  and  51)  gives  the  mantissas  of 
the  logarithms  of  all  integers  from  100  to  1000,  calculated  to 
four  places  of  decimals. 

95.  To  find  the  logarithm  of  a  number  of  three  figures. 

Look  in  the  column  headed  "  No."  for  the  first  two  signifi- 
cant figures  of  the  given  number. 

Then  the  required  mantissa  will  be  found  in  the  correspond- 
ing horizontal  line,  in  the  vertical  column  headed  by  the  third 
figure  of  the  number. 

Finally,  prefix  the  characteristic  in  accordance  with  the 
rules  of  §§  82  and  83. 

For  example,  log  108  =  2.2253  ; 

log  .344  -  9.5300  -  10  ;  etc. 

For  a  number  consisting  of  one  or  two  significant  figures, 
1  he  column  headed  0  may  be  used. 

Thus,  let  it  be  required  to  find  log  88  and  log  9. 

\\y  §  (,)2,  log  83  has  the  same  mantissa  as  log  830j  and  log  9 
the  same  mantissa  as  log  900. 

Hence,  log  83  =  1.9191,  and  log  9  =* 0.9542, 


EXPONENTS  49 

96.  To  find  the  logarithm  of  a  number  of  more  than  three 
figures. 

i.    Required  the  logarithm  of  327.6. 

We  find  from  the  table,     log  327  =  2.5145, 

.  log  328  =  2.5150. 

That  is,  an  increase  of  one  unit  in  the  number  produces  an  increase  of 
.0014  in  the  logarithm. 

Then  an  increase  of  .0  of  a  unit  in  the  number  will  increase  the 
logarithm  by  .6  x  .0014,  or  .0008  to  the  nearest  fourth  decimal  place. 

Whence,  log  327.6  =  2.5145  +  .0008  =  2.£153. 

In  rinding  the  logarithm  of  a  number,  the  difference  between  the  next 
less  and  next  greater  mantissae  is  called  the  tabular  difference  ;  thus,  in 
Ex.  1,  the  tabular  difference  is  .0014. 

The  subtraction  may  be  performed  mentally. 

The  following  rule  is  derived  from  the  above : 

Find  from  the  table  the  mantissa  of  the  first  three 
significant  figures,  and  the  tabular  difference. 

Multiply  the  latter  by  the  remaining  figures  of  the 
number,  with  a  decimal  point  before  them. 

Add  the  result  to  the  mantissa  of  the  first  three 
figures,  and  prefix  the  proper  characteristic. 

In  finding  the  correction  to  the  nearest  units'  figure,  the  decimal  por- 
tion should  be  omitted,  provided  that  if  it  is  .5,  or  greater  than  .5,  the 
units'  figure  is  increased  by  1 ;  thus,  13.26  would  be  taken  as  13,  30.5  as 
31,  and  22.803  as  23. 

2.   Find  the  logarithm  of  .021508. 

Mantissa  215  =  .3324  Tab.  diff.  =     21 

2  .08 

.3326  Correction  =  1.68  =  2,  nearly. 
The  result  is  8.3326  -  10. 

EXERCISE  16 
Find  the  logarithms  of  the  following : 

i.   64.  5.   1079.  9.    .00005023.  13.  7.3165. 

2.  3.7.  6.    .6757.  10.   .0002625.  14.  .019608. 

3.  982.  7.    .09496.        11.   31.393.  15.  810.39. 

4.  .798.         8.   4.288.  12.   48387.  16.  .0025446. 


50 


ALGKIJUA 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

IO 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294- 

0334 

o374 

ii 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

°755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1 106 

13 

1 139 

"73 

1206 

1239 

1271 

J303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

J553 

1584 

1614 

1644 

1673 

'703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

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2122 

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2227 

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17 

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18 

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2577 

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2765 

19 

2788 

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2989 

20 

3010 

3032 

3«54 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

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3365 

3385 

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22 

3424 

3444 

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3483 

3502 

3522 

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356° 

3579 

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3766 

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3874 

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3927 

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25 

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4014 

4031 

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4065 

4082 

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4116 

4133 

26 

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4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

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4378 

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4409 

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28 

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4502 

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29 

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34 

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5328 

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5658 

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5694 

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5866 

5877 

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5899 

39 

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5933 

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40 

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6075 

6085 

6096 

6107 

6117 

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6128 

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6149 

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6201 

6212 

6222 

42 

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6243 

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6263 

6274 

6284 

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631  | 

6325 

43 

6335 

6345 

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6365 

6375 

6385 

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6474 

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6503 

6513 

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6618 

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6628 

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6065 

6675 

6684 

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6702 

6712 

47 

6721 

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6830 

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7101 

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52 

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No. 

0 

1 

2  !  3 

4 

5 

6 

•  7 

8   9 

EXPONENTS 


51 


No. 

0 

1 

2 

3   4 

5 

6 

7   8 

9 

55 

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7435 

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56 

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7589 

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7627 

58 

7634 

7642 

7649 

7657 

7664 

7072 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

773i 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7S10 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

793i 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

88S2 

8S87 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9H3 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

93°9 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

935° 

9355 

9360 

9365 

937° 

9375 

9380 

9385 

939o 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

945° 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

95°4 

9509 

95J3 

95l8 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

958i 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

97°3 

9708 

97J3 

9717 

9722 

9727 

94 

9731 

9736 

974i 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

99<T3 

9908 

98 

9912 

9917 

9921 

9926 

993o 

9934 

9939 

9943 

9948 

9952 

99 

No. 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 
9 

0 

1 

2 

3 

4 

5 

6 

7 

8 

52  ALGEBRA 

97.    To  find  the  number  corresponding  to  a  logarithm, 

i.    Required  the  number  whose  logarithm  is  i.0571. 

Find  in  the  table  the  mantissa  6671* 

In  the  corresponding  line,  in  the  column  headed  "No.,"  we  find  45, 
the  first  two  figures  of  the  required  number,  and  at  the  head  of  the 
column  we  find  4,  the  third  figure. 

Since  the  characteristic  is  1,  there  must  be  two  places  to  the  left  of  the 
decimal  point  (§  82). 

Hence,  the  number  corresponding  to  1.6571  is  45.4. 

2.  Required  the  number  whose  logarithm  is  2.3934. 

We  find  in  the  table  the  mantissa  3927  and  3945. 

The  numbers  corresponding  to  the  logarithms  2.3927  and  2.3945  are 
247  and  248,  respectively. 

That  is,  an  increase  of  .0018  in  the  mantissa  produces  an  increase  of 
one  unit  in  the  number  corresponding. 

Then,  an  increase  of  .0007  in  the  mantissa  will  increase  the  number  by 
T73  of  a  unit,  or  .4,  nearly. 

Hence,  the  number  corresponding 'is  247  +  .4,  or  247.4. 

The  following  rule  is  derived  from  the  above : 

Find  from  the  table  the  next  less  mantissa,  the  three 
figures  corresponding',  and  the  tabular  difference. 

Subtract  the  next  less  from  the  given  mantissa,  and 
divide  the  remainder  by  the  tabular  difference. 

Annex  the  quotient  to  the  first  three  figures  of  the 
number,  and  point  off  the  result. 

The  rules  for  pointing  off  are  the  reverse  of  those  of  §§  82  and  83  : 
I.    If  —  10  is  not  written  after  the  mantissa,  add  1  to  the  characteristic, 
giving  the  number  of  places  to  the  left  of  the  decimal  point. 

II.  If  —  10  is  written  after  the  mantissa,  subtract  the  positive  part  of 
the  characteristic  from  9,  giving  the  number  of  rijihcrs  to  be  placed  between 
the  decimal  point  and  first  significant  figure. 

3.  Find  the  number  whose  logarithm  is  8.5265  —  10. 

52(>5 
Next  less  mant.  sa  5203  ;  liunres  corresponding,  880, 
Tab.  diff.  13)2. ()()(. IT,  =  .2,  nearly. 
1  3 
70 


EXPONENTS  53 

By  the  above  rule,  there  will  be  one  otpber  to  be  placed  between  the 
decimal  point  and  first  significant  figure  ;  the  result  is  .03362. 

The  correction  can  usually  be  depended  upon  to  only  one  decimal 
place ;  the  division  should  be  carried  to  two  places  to  determine  the  last 
figure  accurately. 

EXERCISE  17 

Find  the  numbers  corresponding  to  the  following  logarithms  : 


I. 

0.S189. 

6. 

8.7954  - 10. 

ii. 

1.3019. 

2. 

7.6064  - 

-10. 

7- 

6.5993  -  10. 

12. 

4.2527  - 

-10. 

3- 

1.8767. 

8. 

9.9437  -  10. 

*3- 

2.0159. 

4- 

2.6760. 

9- 

0.7781. 

14- 

3.7264  - 

10. 

5- 

3.9826. 

10. 

5.4571  -  10. 

15- 

4.4929. 

APPLICATIONS 

98.  The  approximate  value  of  a  number  in  which  the  opera- 
tions indicated  involve  only  multiplication,  division,  involu- 
tion, or  evolution  may  be  conveniently  found  by  logarithms. 

The  utility  of  the  process  consists  in  the  fact  that  addition 
takes  the  place  of  multiplication,  subtraction  of  division, 
multiplication  of  involution,  and  division  of  evolution. 

i.    Find  the  value  of  .0631  x  7.208  x  .51272. 

By  §  87,  log  (.0631  x  7.208 x  .51272) 

s±  log  .0031  +  log  7.208  +  log  .51272. 

log   .0631  =    8.8000  -  10 

log   7.208=    0.8578 

log.  51272  =    9.7099-10 
Adding,       log  of  result  =  19.3677  -  20  =  9.3677  -  10.     (See  Note  1.) 
Number  corresponding  to  9.3677  -  10  =  .2332. 

Note  1:  If  the  sum  is  a  negative  logarithm,  it  should  be  written  in 
such  a  form  that  the  negative  portion  of  the  characteristic  may  \&  —  10. 

Thus,  19.3677  -  20  is  written  9.3677  -  10. 

(In  computations  with  four-place  logarithms,  the  result  cannot  usually 
be  depended  upon  to  more  than  four  significant  figures.) 


54  ALGEBRA 

2.  Find  the  value  of      ■  '  * 

7984 

By  §  88,  log  ^  =  log  33(5.8  -  log  7984. 

log 336.8  =  12.5273  -10 
log  7984  =    3.0022 
Subtracting,  log  of  result  =    8i>251  -  10  (See  Note  2.) 

Number  corresponding  =  .04218. 

Note  2 :  To  subtract  a  greater  logarithm  from  a  less,  or  a  negative 
logarithm  from  a  positive,  increase  the  characteristic  of  the  minuend  by 
10,  writing  —  10  after  the  mantissa  to  compensate. 

Thus,  to  subtract  3.9022  from  2.5273,  write  the  minuend  in  the  form 
12.5273  -  10  ;  subtracting  3.9022  from  this,  the  result  is  8.6251  —  10. 

3.  Find  the  value  of  (.07396)5. 

By  §  89,  log  (.07396)5  -    6  X,  log  .07396. 

log  .07396  =    8.8690  -  10 

» 5^       ' 

44.3450  -  50 
=    4.3450  -  10  =  log  .000002213. 

4.  Find  the  value  of  ^.035063. 

By  §  90,  log  ^035063  =  1  log  .035063. 

log  .035063  =  8.5449  -  10 

3)28.5449  -  30  (See  Note  3.) 

9.5150  -  10  =  log  .3224. 

Note  3 ;  To  divide  a  negative  logarithm,  write  it  in  such  a  form  that 
the  negative  portion  of  the  characteristic  may  be  exactly  divisible  by  the 
divisor,*  with  —  10  as  the  quotient. 

Tims,  to  divide  8.5449  —  10  by  3,  we  write  the  logarithm  in  the  form 
28.5449  -  30  ;  dividing  this  by  3,  the  quotient  is  9.5150  -  10. 

EXERCISE  18 

A  negative  number  has  no  common  logarithm  (§  78)  ;  if  such  numbers; 
occur  in  computation,  they  may  be  treated  ;is  if  they  were  positive,  and 
the  sign  of  the  result  determined  irrespective  ol  the  logarithmic  work. 

Thus,  in  Ex.  3  of  the  following  set,  to  tind  the  value  of  (  -96.86)  X&S918 
we  find  the  value  pi  96.86  x  3.8918,  and  put  a  —  sign  before  the  result. 


BXPONEKT8  55 

Find  by  logarithms  the  values  of  the  following : 
1.4.253x7.104.  4    54.029  X  (-.0081487). 

2.  6823.2  x  .1634.  5.    .040764  x  .12896. 

3.  (- 95.86)  x  3.3918.  6.    (-285.46)  x  (-.00070682). 

„     5978  -38.19 

7-   9J62*  I0-    10792'  '*    (8«.08)«. 


8.    21^58.  fl.      670.48  .  ^    C09437)<. 
45057                              -5382.3 

.06405  .000007913  ,o*okm 

r»  I2. k.    (3.625V. 


.002037  .00082375 

Arithmetical  Complement 

99.  The  Arithmetical  Complement  of  the  logarithm  of  a  num- 
ber, or,  briefly,  the  Cologarithm  of  the  number,  is  the  logarithm 
of  the  reciprocal  of  that  number. 

Thus,  colog  409  =  log  ^  =  log  1  -  log  409. 

log  1  =  10.         - 10     (See  Ex.  2,  §  98.) 
log  409=    2.6117 
.-.  colog  409=   7.3883-10. 

Again,  colog  .067  =  log-—-  =  log  1  —  log  .067 

.067 

log  1  =  10.  -  10 

log  .067=   8.8261-10 
•\  colog  .067=    1.1739. 

It  follows  from  the  above  that  the  cologarithm,  of  a  number 
may  be  found  by  subtracting  its  logarithm  from  10  —  10.  _ 

The  cologarithm  may  be  found  by  subtracting  the  last  significant  figure 
of  the  logarithm  from  10  and  each  of  the  others  from  0,  —  10  being  written 
after  the  result  in  the  case  of  a  positive  logarithm. 


56  ALGEBRA 

51384 


Ex.     Find  the  value  of 


8.708  x  .0946 

log       -51384        =  log  L 51384  x-Lxi) 
b8.708  x  .01)40  V  ^-<()«      .0M6/ 


:log  .51384  -flog— —  +  log-  * 


8.708  .0940 

=  log  .51384  +  colog  8.708  +  colog  .0946. 
log  .51384  =  9.7109 -10 
colog  8.708  =  9.0001  -10 
colog  .0946  =  1.0241 

9.7951  -10  =  log  .6239. 

It  is  evident  from  the  above  example  that,  to  find  the  loga- 
rithm of  a  fraction  whose  terms  are  the  products  of  factors,  we 
add  together  the  logarithms  of  the  factors  of  the  numerator,  and 
the  cologarithms  of  the  factors  of  the  denominator. 

The  value  of  the  above  fraction  may  be  found  without  using  cologa- 
rithms, by  the  following  formula : 

log ^M =  iog  .51384  -log(8. 709  x  .0946) 

.    8. 709  x.  0946         8  5V  J 

e=  log  .51384  -  (log  8.709  +  log  .0946). 

The  advantage  in  the  use  of  cologarithms  is  that  the  written  work  of 
computation  is  exhibited  in  a  more  compact  form. 

MISCELLANEOUS   EXAMPLES 

2^5 


100.    i.    Find  the  value  of 


3* 


log^?  =  log  2  +  log  i/6  +  colog  3*     (§  99) 
35 

=  log  2  +  J  log  5  +  j  colog  3. 


log  2=    .3010 
IOg6=    .6990;  -3=    .2330 

colog  3  =  9.5229  -  10  ;   x  9  =  9.(1024  -  10 


.1864  =!(»-!.: 


FACTORS  57 

2.    Find  the  value  of  J1/--03296. 
yi    7.962 

^  •fiBW  =  llog«  =  1  (log  .03296  -  log  7.902). 
°  *  7.902       3        7.962       3  V  b  y 

log  .03296  =  8.5180-10 

log    7.962  =  0.9010 

3)27.6170-30 

9.2057- 10  =  log. 1606. 

The  result  is  —  .1606. 

EXERCISE  19 

Find  by  logarithms  the  values  of  the  following: 

2078.5  x  .05834  (- .076917)  x  26.3 

f'       .3583  x  34o  3'  .5478  x  (-  3120.7)  * 

(J  6.08)  x. 1304  a       .8102  x  (-  6.225)  , 

'"  (-  0721)  x(-  17.976)' 

15.  V6xvl6x^2. 

6    f-  _^18_\* 
1  '    V     8.7  x  .0603J 

•y/  .008546 
-\/.0(>03867 

8    (--14582)*, 
^/^00l'  *?'   7^1000  -(.72346)* 


17 


IV.   FACTORS 

101.  An  irrational  number  is  a  numerical  expression  involv- 
ing surds;  as  -^3,  or  2  +  V5  (§  70). 

102.  A  rational  and  integral  expression  is  resolved  into  its 
prime  factors  when  further  factoring  would  produce  irrational 
factors. 


58  ALGEBRA 

103.   In  the  First  Course  we  considered  the  following  eight 
types  of  factorable  numbers: 

TYPE  FORMS 

I.   a2-b*  =(a  +  b)(a-b). 

II.  a2  +  2a6  +  62^(«  +  6)(a+6), 
a1  -2ab  +  b2  =  (a  -  b)  (a-b). 

III.  as2  +  ate  +  6. 

IV.  aas'2  +  foe  +  c. 

V.  x4  +  ax2y*  +  y*. 

VI.  a3  +  &3=(a  +  &)(a2-a&  +  &2), 
a3  -b*=(a-  b)(a2  +  ab  +  b*). 

VII.   an-bn, 

an  +  6n. 
VIII.  aa?  +  a?/  -h  as  =  a(»  +  y  +  «). 

Of  these  types,  IV  is  more   readily  factored  by  means  of 
VIII  as  follows: 

Ex.    Factor  6  x2  -  7  x  -  20. 

Multiply  —  20  by  6  (the  coefficient  of  x2).     Factor  —  120  so  that  the 

sum  of  the  factors  is  —  7  (the  coefficient  of  x).     These  factors  are  —15,  8. 

Then  write  a    9  0ft      .    ,      1K      ,   Q         **% 

6  x2  —  i  x  —  20  =  6  x2  —  15  x  +  8  x  —  20. 

Group  by  Type  VIII,  s=  3  jc(2  X  -  5)  +  4(2x  -  5), 

whence,  6 x2  -  7  z  -  20  =  (2  x  -  5)  (3  x  +  4) . 

Type  VI  may  be  placed  under  Type  VII. 

EXERCISE   20 

Factor : 

i.  3a?-x-10.  5.   a**  +  4. 

2.  4a2  +  12a +  9.  6.  3s +  8. 

3.  a3-?/3.  7.   a2  +  962-4c2  +  6a&. 

4.  a3  4- a2 -2a -2.  8.  a^+2^+2/2+8(x+^)4-16. 


FACTORS  59 


9.  b*wS-4>+1.  18.  #+d+aM-« 

10.  6a2-17a  +  12.  19.  ar3  +  3#2  +  3;r  +  1. 

11.  9a4-13;r2  +  4.  ;      20.  9ra2-36ran. 

12.  .^  +  7^-8.    •  21.  a»-a*  +  a*-l 

13.  (a.-6)2-2(a-o)-35.        22.  Ga26  -4a2  +  15afc  -  10a. 

14.  m2  +  (a  —  Z>)  m  —  a&.  23.  a"2  —  32. 

15.  m*-l.  24.  9  or4 +  12  or2 +  4. 

16.  (2a-36)2-(a-&)2.  25.  9a2-  30 ab  +  25//-  4  c2. 

17.  p^-f^r1 4- 12.  26.  9a2-25c2  +  62  +  6aZ>. 

27.  36  a4 -61  a2 +  25. 

28.  (3a-&)2-6a(3a-&)  +  27a-96. 

29.  2  a"6  +  250.  32.  a6  +  i/\ 

30.  (7x  +  2)  +  3V7x  +  2  +  2.     33.   16^  +  14^-15. 

31.  m(2a?-3)-4m2a?2  +  9m2.     34.   25(?m +  3)2+10(m  +  3)  +  l. 

35.  8(2a-56)-1-12(2a-56)"M-4. 

36.  143  A:2 -103  A: +  14. 

37.  Var  +  4a-6  +  2a2-l  +  4(2#-3). 

38.  g*  +  g2t2  +  t\  42.  *P+$. 

39.  x*  +  a2x  —  a3  —  az2.  43.  a?4  —  13  x2  +  4. 

40.  c-3d-18-9d  +  2c~*.  44-  9  ey- 16  e/3. 

41.  r4-20r2  +  99.  45.  304  v2  +  25  v  -  6. 

46.  (a>  +  l)'  +  2(a>  +  l)*  +  l. 

47.  &(*+ff  +  H*+fff(*f fl- 
ip. a6 -64.  50-  25a4  +  a2,+  l. 
49.   6  or4  —  41  x~hfr  —  7  y.             51.   4+ a3  —  a2  —  4  a. 

52.  m4  —  1  +  ra  —  m3. 

53.  3(^  +  l)+5(x-2-l)  +  (.r  +  l)2. 

54.  ear^  +  lSar1-^.  57-  ;>2-0- 

55.  a*+fti  58.   4a?*+4-4a*+2  +  l. 

56.  **-#V.  59-   a2x-9^  +  2a--18. 


60  ALGEBRA 

60.  4/>V  +  20pr/-lG/A/-80^/. 

61.  (x2  -  2  x  + 1)  -  (a  + 1)2.        65.   a5-  27  a2  +  243  -9  a3. 

62.  52m  — 10m2  — 10.  66.   22m -\- A  x  •  2m  —  21  x2. 

63.  s*»-j*  67.  a  +  2Va&  +  &. 

64.  a8 -256.  68.   (2  a  -  3&)2-  (3  a-  2  &)2. 

69.  a2  +  2a&-f-c2-2&c-2ac+62. 

70.  27m3-54m2-f-36m-8.      72.   9  a2- 6a -4  62- 45. 

71.  ar5  +  a?4  +  oj8 4- #24^  +  l.         73.   1  +  2ab  -  a4-  a2b2  -  64. 

74.  am2  —  ms  4  2  am 71  4-  an2  4-  2  m2n  —  mn2. 
75-  2/2  +  rz-2?/-z4-l. 

FACTOR  THEOREM 

104.    The  Remainder  Theorem. 

Let  it  be  required  to  divide  px2  4-  qx  4-  r  by  a;  —  a. 

pas2  +  #x  +  r  I  x  —  a 

px*  -  apx      \px  +  (ap  +  g) 

(ap  +  g)a 

(ap  4  q)x  —  pa2  —  ga 

pa'1  +  qa  +  r,  Remainder. 

We  observe  that  the  final  remainder, 
pa2  +  qa-\-r, 
is  the  same  as  the  dividend  with  a  substituted  in  place  of  x ; 
this  exemplifies  the  following  law  : 

If  any  polynomial,  involving  xt  be  divided  by  x  -  a,  the 
remainder  of  the  division  equals  the  result  obtained  by 
substituting  a  for  x  in  the  given  polynomial. 

This  is  called  The  Remainder  Theorem. 

To  prove  the  theorem,  let 

pxn  4-  qxn~l  4-  •••  +r#4s 
be  any  polynomial  involving  x. 

Let  the  division  of  the  polynomial  by  #  — a  be  (ferried  on 
until  a  remainder  is  obtained  which  does  not  contain  x. 

Let  Q  denote  the  quotient,  and  E  the  remainder. 


FACTORS  61 

Since  the  dividend  equals  the  product   of  the  quotient  and 
divisor,  plus  the  remainder,  we  have 

Q(x  —  a)  +  B'=:pa?  +  qx?-l4-  ...  +  rx  +  s. 

Putting  x  equal  to  a,  into  the  above  equation,  we  have, 
11  =2Mn  +  qan~l  -j-  ...  +ra  +  s. 

105.  The  Factor  Theorem. 

If  any  polynomial,  involving  x,  becomes  zero  when  as 
is  put  equal  to  a,  the  polynomial  has  a?  —  a  as  a  factor. 

For,  by  §  104,  if  the  polynomial  is  divided  by  x  —  a,  the 
remainder  is  zero. 

106.  Examples. 

i.   Find  whether  x  —  2  is  a  factor  of  ar3  —  5  ic2  -f  8. 

Substituting  2  for  sc,  the  expression  xs  —  5  x2  +  8  becomes 

23  _  5  .  22  +  8,  or  -  4. 

Then,  by  §  104,  if  r3  —  5  x2  +  8  be  divided  by  x  —  2,  the  remainder  is 
—  4 ;  and  sc  —  2  is  not  a  factor. 

2.    Find  whether  m  +  wisa  factor  of 

m4  —  4  m*ra  -f-  3  m2a2  +  5  mn3  —  2  ?*4.  (1) 

Putting  m  =  —  n,  the  expression  becomes 

7i4  4-  4  n4  -1-  2  ra4  —  5  n*  —  2  n4,  or  0. 

Then,  by  §  104,  if  the  expression  (1)  be   divided  by  ra  4-  ?i,  the  re- 
mainder is  0  ;  and  m  +  n  is  a  factor. 

•    3.    Prove  that  a  is  a  factor  of 

(94-$,+  c)(afr  +  &c  +  ca)  -  (a  +  &)(&  +  c)(«  +  «■)• 
Putting  a  ±=  0,  the  expression  becomes 

(b  4-  c)frc  -  $(ft  4-  c)c,  or  0. 
Then,  by  §  104,  a  —  0,  01  a,  is  a  factor  of  the  expression. 


62  ALGEBRA 

4.   Factor  x*-3x2-  Ux-  8. 

The  positive  and  negative  integral  factors  of  8  are  1,  2,  4,  8,  —  1,  —  2, 
-  4,  and  -  8. 

It  is  best  to  try  the  numbers  in  their  order  of  absolute  magnitude. 

If  x  =  1,  the  expression  becomes  1—3—14  —  8. 

If  x  =  —  1,  the  expression  becomes  —  1  —  3  +  14—8. 

If  x  =  2,  the  expression  becomes  8—12  —  28  —  8. 

If  x  =—  2,  the  expression  becomes  —8—12+28  —  8,  or  0. 

This  shows  that  x  +  2  is  a  factor. 

Dividing  the  expression  by  x  +  2,  the  quotient  is  x2  —  5  x  —  4. 

Then,  x3  -  3  x2  -  14  a  -  8  =  (x  +  2)02  -  5  x  -  4). 

EXERCISE  21 

Factor  the  following : 

1.  as  +  8.  9.  an-6n. 

2.  m5-f-?r\  10.  2ar3-f5a;2  —  #  —  6. 

3.  a;6- 729.  11.  a4-ar*  +  2a;2-4. 

4.  ar3  + 5a;2-8a;  +  2.  -    12.  5a3-18a-4. 

5.  m3- 11  ?7i  —  10.          •  13.  ar3  + a,-2 +-7  a; +  18. 

6.  a4  —  a3  +  3  a  —  14.  14.  m3  —  5  m2  —  36. 

7.  c3_2c2-9.  15.  A;4-5A;2  +  3A;-2. 

8.  a;4 -625. 

Find  without  actual  division  : 

16.  Whether  p  —  1  is  a  factor  of  p3  -+  3p2  —  4. 

17.  Whether  a;  -f-  2  is  a  factor  of  xA  -f  3  x3  —  4  a,\ 

18.  Whether  a?  +  1  is  a  factor  of  2  a?  +  6  x2  -  3  a?  +  4. 

19.  Whether  m  —  3  is  a  factor  of  m3  —  4  m  —  15. 

20.  Whether  a  —  5  is  a  factor  of  a3  —  3  a2  —  5  a  —  25. 

21.  Whether  c  -  2  is  a  factor  of  3  c3  —  9  c2  +  5  c  +  2. 

22.  Whether  a  is  a  factor  of  a(b  —  c)  -f  b  (c  —  a)  +  c(a  —  b). 

23.  Whether  c  is  a  factor  of  a (b  —  c)  +  b (c  —  a)  +  c(a  —  b). 

24.  Whether  x  +- 1/  is  a  factor  of  a?  (2  a?  -f-  3  ?/)  —  y  (3  a?  +-  2  // ). 

25.  Whether  b  is  a  factor  of  a2  (b  -  c)2  +  62  (c  -  a)2  +  c2  (a  -  b)2. 


FACTORS  63 

HORNER'S   SYNTHETIC   DIVISION 

107.  The  method  of  synthetic  division,  or  as  it  is  sometimes 
known,  the  method  of  detached  coefficients,  greatly  abridges 
the  work  of  division,  especially  where  binomial  divisors  are 
concerned. 

108.  Divide  ar3  -  11  x2  +  36  x  -  36  by  x  -  3. 

Writing  dividend  and  divisor  with  coefficients  only, 

1-3 

1-8  +  12     Quotient. 


1-11  +  36- 

-36 

1-    3 

-    8 

-    8  +  24 

+  12- 

-36 

+  12- 

-36 

Since  the  first  term  of  each  partial  product  is  merely  a  repetition  of 
the  term  immediately  above,  it  may  be  omitted. 

We  may  also  change  the  sign  of  the  second  term  of  the  divisor  if  the 
partial  product  is  added  instead  of  subtracted. 
We  then  have 

1  _  ii  +  36  -  36  II  +3 
1+3  11-8  +  12 

-8 

-24 

•       +12 

+  36 

Raise  the  numbers  —  24,  36  now  in  the  oblique  column  and  the  work 

stands  : 

1-  11+36-361  +  3 

4-    3-24  +  36 

-       +12 

The  quotient  is  x2  —  8  x  +  12. 

If  the  last  remainder  is  zero,  x  minus  the  divisor  is  a  factor  of  the 

expression. 


64  •  ALGEBRA 

EXERCISE  22 

Divide  the  following  by  synthetic  division: 

i .  2  x^  —  7  x2  +-  x  +  10  by  x  —  2. 

2.  3  a4  —  a3  —  5  a2  +  6  a  +  7  by  a  -h  1. 

3.  a4-lla3  +  29a2-9a+-14by  a-7. 

4.  4  m3  — 17  m2?i  -f-  13  mn2  -f-  0  ft*  by  m  —  3  fl. 

5.  3a5  +  lla4  _43^-4a;2-f  lla-6  by  »  +  6. 

6.  8^4-35?;3  +  7v2  +  22v-8by  v-4. 

109.    Divide  ar3  —  11  a;2  +  36  x  —  36  by  x  —  5,  and  by  a  —  7. 

1  -  11  +  36  -  36  |6  1  -  11  +  36  -  36  [7_ 

■4.    5-30  4-30  -f   7-28  +  56 

—   6+6—6     Remainder  —    4  +    8  +  20     Remainder 

(Quotient)  >    (Quotient) 

A  factor  lies  between  x  —  5  and  x—7.     It  is  found  to  be  x  —  6. 

Then  if  in  dividing  by  a  binomial  a  remainder  occurs,  and  if 
the  remainders  arising  from  successive  division  by  two  binomi- 
als are  of  opposite  sign,  a  factor  #—  a  lies  between  these  two 
binomials. 

EXERCISE   23 

i.    Locate  the  root  between  2  and  4  of  x3  —  17  x  -f  24  =  0. 
Locate  roots  of  the  following : 

2.  a3  +  10a2  +  17a-2S  =  0. 

3.  a4  +  3tt3-10a2  +  3a  +  ir>  =  0. 

4.  x5  —  8xi-7xi  +  Mx2-5x  +  4i)  =  0. 

5.  3ar3-26#2  +  60<»-72  =  0. 

6.  mA  -  2  m3  -  19  m2  +  12  m  +  40  =  0. 


FACTORS  65 

SOLUTIONS 

110.    If  the  product  of  abc  •••  to  n  factors  =  0,  at  least  one 

of  the  factors  must  be  zero. 

Ex.  1.    Let  (x -2)(x-  3)(x  +  4)  =  0. 

Then  x  —  2,  x  —  3,  or  x  +  4  must  equal  zero. 

The  equation  is  satisfied  by  the  root  obtained  by  putting  any  one  of 
the  factors  equal  to  0.  Hence,  x  =  2,  3,  or  —  4  are  the  solutions  of  the 
equation. 

Ex.   2.     Solve  52*- 5* -12  =  0.  .  (1) 

(5*-4)(5*  +  3)=0.  (2) 

Whence,  5*  -  4  =  0,     5*  =  4,  (3) 

and  5*  +  3  =  0,     5»  =  -  3.  (4) 

To  solve  (3)  and  (4),  take  the  logarithms  of  each  member  of  the 
equations : 

From  (3)  zlog5  =  log  4  (§  89) ,  (5) 

and  ^log4  =  10020  =  602# 

log5       .6990      699  v  J 

From  (4)  x  log  5  =  —  log  3. 

Ex.  3.    Solve  the  equation  .2*  =  3. 

Taking  the  logarithms  of  both  members,  xlog.2  —  log  3. 

Then  x  =  ^^=        A1U        =  -^2L  =  _  .6285+. 

log. 2       9.3010-10       -.699 

An  equation  of  the  form  ax  =  b  may  be  solved  by  inspection 
if  b  can  be  expressed  as  an  exact  power  of  a. 

Ex.  4.    Solve  the  equation  16x  =  128. 

We  may  write  the  equation  (24)*  =  27,  or  24*  =  27. 
Then,  by  inspection,  4  x  =  7  ;  and  x  =  J. 

(If  the  equation  were  16x  =  — ,  we  could  write  it  (24)*  =  —  =  2"7  ; 
v  4  128  27 

then  4x  would  equal  —  7,  and  a;  =—  {.) 


Xx-A 


66  ALGEBRA 

EXERCISE   24 

Solve  the  following  equations  : 

i.  13*  =  8.  4.  .005038*  =  816.3.  7.  .2*+5  =  .5* 

2.  .06*  =  .9.  5.  34*-1  =  42*+3.  8.  16*  =  32. 

3.  9.347*  =  .0625.        6.  73*+2  =  .8*.  9.  32*  =  ^ . 

10.  (TV)^  =  8.         11.  ay  =  Jj.        12.  .042*-  5  (.04)*  -24  =  0. 

13.  23*  +  7.22*-  9.2*  -63  =  0. 

14.  3°"-5.32'- 8.3* +  12  =  0. 

15.  II4* -5- II2* +  4  =  0. 

16.  23*+3-6.22*+2  +  ll  •  2*+1-6  =  0. 

17.  .54*  -  2(.5)3*  -  16  (.5)2*  +  2  (.5)*  +  15  =  0. 

18.  23*  -  10- 22*- 71 -2* -60  =  0. 

19.  a?-x2-9x  +  9  =  Q. 

20.  ar  +  (5c  +  2c7)£  +  10ccZ  =  0. 

COMMON   FACTORS   AND   MULTIPLES 

111.  A  Common  Factor  of  two  or  more  expressions  is  a  factor 
of  each  of  them. 

112.  The  Highest  Common  Factor  (H.  C.  F.)  of  two  or  more 
expressions  is  their  common  factor  of  highest  degree  (§  23). 

113.  A  Common  Multiple  of  two  or  more  expressions  is  an 
expression  which  is  exactly  divisible  by  each  of  them. 

114.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
expressions  is  their  common  multiple  of  lowest  degree. 

Ex.  1.    Find  the  H.  C.  F.  of  a2  +  2  a  -  3  and  1  -  a3. 

a'2+2a-3=  (a-l)(a  4-3). 

1-0,3  =  (1_«)(1  +  rt  +  a2). 

The  factors  of  the  first  expression  can  be  put  in  the  form 

-  (1  -a)(8  +  a). 
Hence,  the  H.  C.  F.  is  1  -  a. 


FACTORS  67 

Ex.  2.    Eequired  the  L.  C.  M.  of 

x2  —  5x  +  6,  x2  —  4x  +  4,  and  Xs  —  9  x. 

•    *2-5x+6  =  (£-3)(z-2). 

x2-4x+4=:  0-2)2. 

x*  -  9  x  =  x(x  +  S)(x  -  3). 

It  is  evident  by  inspection  that  the  L.  C.  M.  of  these  expressions  is 
z(z-2)2(£  +  3)(jc-3). 

115.  "When  the  polynomials  cannot  be  readily  factored  by 
inspection,  the  H.  C.  F.  and  L.  C.  M.  may  be  found  by  the  fol- 
lowing method. 

The  rule  in  Arithmetic  for  the  H.  C.  F.  of  two  numbers  is : 

Divide  the  greater  number  by  the  less. 

If  there  be  a  remainder,  divide  the  divisor  by  it;  and 
continue  thus  to  make  the  remainder  the  divisor,  and 
the  preceding  divisor  the  dividend,  until  there  is  no 
remainder. 

The  last  divisor  is  the  H.  C.  F.  required. 

Thus,  let  it  be  required  to  find  the  H.  C.  F.  of  169  and  546. 

169)546(3 
507 
39)169(4 
156 

13)39(3* 
39 
Then  13  is  the  H.  C.  F.  required. 

116.  We  will  now  prove  that  a  rule  similar  to  that  of  §  115 
holds  for  the  H.  C.  F.  of  two  algebraic  expressions. 

Let  A  and  B  be  two  polynomials,  arranged  according  to  the 
descending  powers  of  some  common  letter. 

Let  the  exponent  of  this  letter  in  the  first  term  of  A  be 
equal  to,  or  greater  than,  its  exponent  in  the  first  term  of  B. 

Suppose  that  B  is  contained  in  A  p  times,  with  a  remainder 
C;  that  C  is  contained  in  B  q  times,  with  a  remainder  D;  and 
that  D  is  contained  in  C  r  times,  with  no  remainder. 


68  ALGEBRA 

To  prove  that  D  is  the  H.  C.  F.  of  A  and  B. 

The  operation  of  division  is  shown  as  follows  : 

B)A(p 
pB 
C)B(q 
qC_ 
D)C(r 
rD 
0 

We  will  first  prove  that  D  is  a  common  factor  of  A  and  B. 

Since  the  minuend  is  equal  to  the  subtrahend  plus  the  remainder 
(F-C"§4°)'  A^pB+C,  (1) 

B  =  qC+D,  (2) 

and  C  =  rD. 

Substituting  the  value  of  C  in  (2),  we  obtain 

B  =  qrD  +  D=  D(qr  +  1).  (8) 

Substituting  the  values  of  B  and  C  in  (1),  we  have, 

A  =pD(qr  +  1)  +rD  =  D(pqr  +p  +  r).  (4) 

From  (3)  and  (4),  D  is  a  common  factor  of  A  and  B. 

We  will  next  prove  that  every  common  factor  of  A  and  B 
is  a  factor  of  D. 

Let  F  be  any  common  factor  of  A  and  B ;  and  let 
A  =  mJP7,  and  B  =  nF. 

From  the  operation  of  division,  we  have 

C  =  A-pB,  (5) 

and  D  =  B-qC.  (6) 

Substituting  the  values  of  ^4  and  7?  in  (5),  we  have 
C  =  mF  —  pnF. 

Substituting  the  values  of  B  and  C  in  (6)  we  have 

D  =  nF—  q(mF  —  pnF)=  F(n  —qm  +pqn  ). 
Whence,  F  is  a  factor  of  D. 


FACTORS  ill) 

Then,  since  every  common  factor  of  A  and  B  is  a  factor  of 
]),  and  since  D  itself  is  a  common  factor  of  .1  and  />,  it  follows 
that  D  is  the  highest  common  factor  of  A  and  B. 

We  then  have  the  following  rule  for  the  H.  C.  F.  of  two 
polynomials,  A  and  B,  arranged  according  to  the  descending 
powers  of  some  common  letter,  the  exponent  of  that  letter  in 
the  first  term  of  A  being  equal  to,  or  greater  than,  its  exponent 
in  the  first  term  of  B: 

Divide  A  by  B. 

If  there  be  a  remainder,  divide  the  divisor  by  it; 
and  continue  thus  to  make  the  remainder  the  divisor, 
and  the  preceding  divisor  the  dividend,  until  there  is 
no  remainder. 

The  last  divisor  is  the  H.  C.  P.  required. 

It  is  important  to  keep  the  work  throughout  in  descending  powers  of 
some  common  letter  ;  and  each  division  should  be  continued  until  the 
exponent  of  this  letter  in  the  first  term  of  the  remainder  is  less  than  its 
exponent  in  the  first  term  of  the  divisor. 

Note  1 :  If  the  terms  of  one  of  the  expressions  have  a  common  factor 
which  is  not  a  common  factor  of  the  terms  of  the  other,  it  may  be  re- 
moved ;  for  it  can  evidently  form  no  part  of  the  highest  common  factor. 

In  like  manner,  we  may  divide  any  remainder  by  a  factor  which  is  not 
a  factor  of  the  preceding  divisor. 

117.    i.   Find  the  H.  C.  F.  of 

Gx2 - 25x  +  14  and  Ox*- 7 x2 - 25 x  + 18. 

6x2-25x  +  14)Gx*-    7a---25x+  18(x  +  3 
6s* -25x2  +  Ux 
18x2-39x 
18s2 -75s +  42 
36x-24 


In  accordance  with  Note  1,  we  divide  this  remainder  by  12,  giving 

x~~2'  3x-2)()x2-25x  +  14(2x-7 

n.i-2-    4x 

-21*  -     # 

-'21  as +  14 

Then,  Sx  —  2  is  the  H.  C.  F.  required. 


70  ALGEBRA 

Note  2 :  If  the  first  term  of  the  dividend,  or  of  any  remainder,  is  not 
divisible  by  the  first  term  of  the  divisor,  it  may  be  made  so  by  multiply- 
ing the  dividend  or  remainder  by  any  term  which  is  not  a  factor  of  the 
divisor. 

2.    Find  the  H.  C.  F.  of 

3  «8  +  a2b  _  2  atf  an(i  4  asb  +  2  a2b2  -  abs  +  ft4. 

We  remove  the  factor  a  from  the  first  expression  and  the  factor  b  from 
the  second  (Note  1),  and  find  the  H.  C.  F.  of 

3  a2  +  ab  -  2  V2  and  4  a3  +  2  a2ft  -  aft2  +  ft3. 

Since  4  a3  is  not  divisible  by  3  a2,  we  multiply  the  second  expression 
by  3  (Note  2). 

4  a3  +  2  a2ft  -     aft2  +     ft3 

3^ 

3  a2  +  a&  _  2  ft2)l2  a3  -f  6  a2ft  -  3  ab2  +  3  ft3(4  a 
12q8  +  4a2ft-8aft2 

2  a2ft  +  5  aft2  +  3  ft3 

Since  2a2ft  is  not  divisible  by  3  a2,  we  multiply  this  remainder  by 
8  (Note  2). 

2  a2b  +    5  ab2  +    3  ft3 
3 


3  a2  +  aft  -  2  ft2)6  a2ft  +  15  aft2  +    9ft3(2ft 
6a2&+    2  aft2-    4  ft3 
13  aft2 +  13  ft3 

We  divide  this  remainder  by  13  b2  (Note  1),  giving  a  +  b. 

a  +  6)3  a2  +     aft  -  2  ft2(3  a  -  2  6 
3  a2  +  3  aft 
-2  aft 
-  2  aft  -  2  ft2 

Then,  a  +  ft  is  the  H.  C.  F.  required. 

Note  8 :  If  the  first  term  of  any  remainder  is  negative,  the  sign  0! 
each  term  of  the  remainder  may  be  changed. 

Note  4:  If  the  given  expressions  have  a  common  factor  which  can 
be  se£n  by  inspection,  remove  it,  and  find  the  H.C.  P,  of  the  resulting 
expressions;  the  result,  multiplied  by  the  common  factor,  will  be  the 
H.  C.  F.  of  the  given  expressions. 


FACTORS  71 

3.   Find  the  H.  C.  F.  of 

2aJ4  +  333-6a2  +  2ajand  ffV^-W  — 2flf  — ft 

Removing  the  common  factor  x  (Note  4),  we  find  the  H.  C.  F.  of 
2x3  +  8 as*  -  8a; -f  2  and  Gx3  +  5x2  -  2x  -  1. 
"  2x3  +  3x2-()x  +  2)6a*  +  5*»-    2x-l(3 
6x3  +  0x2-  18a; +  6 
-4x2  +  16x-  7 

The  first  term  of  this  remainder  being  negative,  we  change  the  sign  of 
each  of  its  terms  (Note  3). 

2x3  +    3x2-      6x+    2 
2 


4x2-  16x  +  7)4x3 -f-    6x2-    12x  +    4(x 
4x3-16x2  +      7x 

22x2-    19x  +    4 
2 


44 x2-    38x+    8(11 
44x2-  176x  +  77 
69)138x-69 
2x-    1 

2x-l)4x2-    16x  +    7(2x-7 
4  x2  —      2  x 

-  14x 

-  14x4-  7 

The  last  divisor  is  2x—  1  ;  multiplying  this  by  x,  the  H.  C.  F.  of  the 
given  expressions  is  x  (2  x  —  1). 

(In  the  above  solution,  we  multiply  2  x3  -f  3x2  —  6x  -b  2  by  2  in  order 
to  make  its  first  term  divisible  by4x2;  and  we  multiply  the  remainder 
22  x2  —  19  x  -h  4  by  2  to  make  its  first  term  divisible  by  4  x2.) 

118.  We  will  now  show  how  to  find  the  L.  C.  M.  ot  two  ex- 
pressions which  cannot  be  readily  factored  by  inspection. 

Let  A  and  B  be  any  two  expressions. 
Let  F  be  their  H.  C.  F.,  and  M  their  L.  C.  M. 
Suppose  that  A  =  aF,  and  B  =  bF. 

Then,  AxB  =  abF2.  (1> 

Since  .Pis  the  H.  C.  F.  of  A  and  #,  a  and  b  have  no  common  factors  ; 
whence  the  L.  C.  M.  of  rti^and  bF  is  abF. 
That  is,  M=abF. 


72  ALGEBRA 

Multiplying  each  of  these  equals  by  F,  we  have 

Fx  M=ahF~.  (2) 

-  From  (1)  and  (2),        A  x  B  =  F  x  M. 

That  is,  the  product  of  two  expressions  is  equal  to  the  product 

of  their  H.  C.  F.  and  L  C.;M. 

Therefore,  to  find  the  L.  C.  M.  of  two  expressions, 
Divide  their  product  by  their  highest  common  factor  ;  or, 
Divide  one  of  the  expressions,  by  their  highest  common  factor, 

and  multiply  the  quotient  by  the  other  expression. 

Ex.   Find  the  L.  C.  M.  of 

6x2-17x2  +  12  and  12a2-4a-21. 

6x2-  17 x  +  12)12 x2-    4x-21(2 
12  x2  -34  x  +  24 
15)30  x-  45 

2x-  3)6x2-17x  +  12(3«-4 
6x2-9x 
-Sx 
-8x  +  12 


Then,  the  H.C.F.  of  the  expressions  is  2x  —  3. 

Dividing  Ox*2  —  17  x  +  12  by  2  x  —  3,  the  quotient  is  3  x  —  4. 

Then,  the  L.  C.  M.  is  (3x- 4)(12x2  -  4x- 21). 

EXERCISE  25 

Find  the  H.  C.  F.  and  L.  C.  M.  of  the  following: 

i.  2a2  +  a-6,    4  a2  -8a  +-3. 

2.  6a;2-17a,'  +  10,    9a?2-14^-8. 

3.  x2-Cyx-27,    x>-2x2-8x  +  2l. 

4.  6^2-31^/  +  182/2,    9 x2  +  15 ^-  14 y\ 

5.  8a2  +  6a-9,    6a3  +  7a2-7a-6. 

6.  4a2-lla-3,    8xA  +  6x*-llx2-23x-5. 

7.  m5 -4m3  +  ^2-4,    m4 — 2 m3  —  m*  +  m  +  2. 

8.  12p2-19pg-21?2,    Y2lr  +  r,lr<i-npq2-(S(f. 

9.  c4  +  7c3+12c2,    c3  +  4c2-9c-36,    S^+lOcr-lSc-L'S. 


FRACTION'S  73 

10.  8  ar3  + 27,   4  a3-  8x2-  9x  +  IS,    2  .r  +  ar°-  11  a- 12. 

ii.  81- it-4,    ic4-4ir3  +  4x2-4x  +  3. 

12.  a4  +  4,   aar  +  2a.c  +  2a,    ar3  +  3a;2  +  4#  +  2. 

13.  16c4  +  8c2  +  81,   4c3  +  4c2  +  c-9. 

14.  a3  +  7a2-9a-63,    a3  +  f>a2-{- 11a +  6. 

15.  (5a-3bf-(a  +  b)\    72a2-  48  a&  +  8&2. 

16.  a3  +  6a2a  +  12aa;2  +  8ar3,   4  a*  +  8  a4x  -  a  V  -  2  aV. 

V.    FRACTIONS 

119.  A  Fraction  is  an  indicated  quotient  written  usually  in 
the  form  — ,  where  a  is  the  dividend,  and  is  called  the  numera- 
tor, and  b  the  divisor,  and  called  the  denominator. 

120.  If  the  same  factor  be  introduced  into,  or  removed  from, 
both  dividend  and  divisor,  the  quotient  is  not  changed.  Upon 
this  principle  depends  the  reduction  of  fractions  to  either 
higher  or  lower  terms.  The  laws  of  sign  for  fractions  are 
those  of  ordinary  division.  The  sign  before  the  fraction  de- 
notes whether  the  quotient  is  to  be  added  or  subtracted. 

REDUCTION  OF  FRACTIONS 

121.  Change  of  sign, 

-+-«_  _  —  cl  _  _  +  a  _  y  ♦—  a 

EXERCISE  26 

Write  each  of  the  following  in  three  other  ways  without 
changing  its  value : 


a 

2' 

2.  S±».   3.      8    • 

7         °      2-x 

2x-l               &x-5 
4"    x  +  2        5-    (x  -3X^+4) 

6       &2_"2 
'    2b2-a?' 

(ix-3y)(y-3x)^ 
(2y  +  x)(x-y) 

74  ALGEBRA 

122.  Reduction  to  Lowest  Terms.  This  is  accomplished  by 
removing  every  factor  common  to  both  numerator  and  denomi- 
nator. If  numerator  and  denominator  are  not  prime  to  each 
other,  it  is  possible  generally  to  factor  them  by  inspection. 
When,  however,  the  factors  cannot  be  readily  seen,  the  method 
of  §  117,  known  as  the  Euclidean  method,  may  be  used. 

EXERCISE  27 
Reduce  the  following  to  lowest  terms : 


27  a8  +  8 

9  x2  +  12  x  +  4 

a8  _  a*b  _  abA  +  h5 

a4-a?b-a2b2  +  ab3 

12z2  +  16zy-3y2 

2-     Z2 "^ M,       _,.•  5- 


10 2* +  22/ -21 2/*  6'    14  as* +  14  as -281 


2a*  +  5ar°-2 

x  +  3 

6a^-7^2  +  5 

x  —  2 

x?  —  x2  —  4  x 

-6 

a?  +  7rf  +  12x  +  10 

16  x2  + 16  x  - 

32 

Simplify  the  following: 

7-    ft'-g«%l^       4<l 


a2  +  5  a  -f  4/ \        a2  —  2  a  + 1 


~     x2  —  y2      y  _  2  x*y 


x*  +  xy      x      xr  -f  ?/2 

15a  +  56      c2-3c-10  .  ac  +  2a-14-7c 

■  x 


c3-125  e2-8c  +  16      2c-8-4a  +  ac 

11  2  .r             4  x* 

x  +  2     x-2  x2  +  ±     x*  +  16 

a-\-b      a  — & 


a  —  b     a  +  b 
"'    «2  +  fr2     a2-"^' 

a2_&2         a2  +  &2 

*  To  simplify  this  and  following  examples  of  Exercise  27,  perform  the 
indicated  operations,  then  reduce  the  resulting  fraction  to  its  lowest  terms. 


12. 


13- 


14. 


15- 


be 


FRACTIONS 
ab 


75 


(a  —  b)(a  —  c)      (c  —  a)(b  —  e)      (c  —  b)(b  —  a) 


4         5a2-7a-6     8a2-16a  +  6 


a_l      6a2-a-12       2a2-5a  +  2 
4 


3a- 


x  —  y 


-j  _2x  —  3y 
3x  —  4:y 


m  —  2  _  4  —  ?n     1  —  m 
m  -f  5      3  —  m     m  —  5 


16.      3a+5- 


i7- 


a  +  2 


-K 


5  a 


a2-10a-24 


+  1 


Sx  +  2y     3x-2y 


Ntf  +  Sf 


1_1 

27arJ-82/3J 


18. 


19. 


4c4-29c2+25  _  /4c2-20c  +  25     6c2  +  llc-l(P 
c6-l  ;   V  3c2  +  3c  +  3  9c2-4       , 


g     s  9fa2  +  5a?-f  6) 


^  +  3^4-3  a?  4-1 


M 

x      y 


4x 


2  +  > 


21. 


1  _i_  4  «ff  +  f         ^  ^  +  K8 
4  a;2 


x  +  4:     x  —  1      a;  +  2       a?2  — a;- 16 


#  +  2      #  —  3      #—5      a:2  —  8  x  + 1 5 

1      a^4-2a;-ll\  9 

af'+5a;-14y  *  ^  +  343* 


76  ALGEBRA 

2         &2  +  3a>  +  2  4  ,  x2  +  l 


24. 


SB  + 1  a?"  —  1  #2  -|-  5  x  +  6         #  12  a; 

(2x2-2xy-2x)(x2-y2)^ 


X 


x  +  y 


cf  +  V  a4_&4  3a2 

■     a*  +  &2     a2b  +  ab2     a4  -  a?b +  d2b2 -ab3+b* 

123.  Under  certain  conditions  a  fraction  may  assume  a  form 
the  value  of  which  is  not  readily  seen.  Such  forms  usually 
occur  in  limiting  values  of  fractions  in  which  the  unknown  or 
unknowns  are  considered  variable. 

124.  A  variable  number,  or  simply  a  variable,  is  a  number 
which  may  assume,  under  the  conditions  imposed  upon  it,  an 
indefinitely  great  number  of  different  values. 

A  constant  is  a  number  which  remains  unchanged  throughout 
the  same  discussion. 

125.  A  limit  of  a  variable  is  a  constant  number,  the  differ- 
ence between  which  and  the  variable  may  be  made  less  than 
any  assigned  number,  however  small. 

Suppose,  for  example,  that  a  point  moves  from  A  towards  B  under  the 
condition  that  it  shall  move,  during  successive  equal  intervals  of  time, 
first   from   A   to    O,    halfway   be- 
tween i  andi?;  then  to  Z>,  half-    f L ?      *      f 

way  between   C  and  B ;    then  to 

E,  halfway  between   D  and  B ;  and  so    on   indefinitely. 

In  this  case,  the  distance  between  the  moving  point  and  B  can  be  made 
less  than  any  assigned  number,  however  small. 

Hence,  the  distance  from  A  to  the  moving  point  is  a  variable  which 
approaches  the  constant  value  A  B  as  a  limit. 

Again,  the  distance  from  the  moving  point  to  B  is  a  variable  which 
approaches  the  limit  0. 

126.  Interpretation  of  j  • 

Consider  the  series  of  fractions  -,  — ,  — '-,  — — ,  •••. 

3'  .3'  .03 '  .003' 


FRACTIONS  77 

Here  each  denominator  after  the  iiist  is  one-tenth  of  the 
preceding  denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  less  than  any  assigned  number, 
however  small,  and  the  value  of  the  fraction  greater  than  any 
assigned  number,  however  great. 

In  other  words, 

If  the  numerator  of  a  fraction  remains  constant,  while 
the  denominator  approaches  the  limit  0,  the  value  of 
the  fraction  increases  without  limit. 

It  is  customary  to  express  this  principle  as  follows : 

a 

The  symbol  go  is  called  Infinity ;  it  simply  stands  for  that  which  is 
greater  than  any  number,  however  great,  and  has  no  fixed  value. 

127.   Interpretation  of  —  • 

00 


Consider  the  series  of  fractions  -,  — ,  — — 


3?  30'  300'  3000'      * 
Here  each  denominator  after  the  first  is  ten  times  the  pre- 
ceding denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  greater  than  any  assigned  number, 
however  great,  and  the  value  of  the  fraction  less  than  any 
assigned  number,  however  small. 
In  other  words, 

If  the  numerator  of  a  fraction  remains  constant,  while 
the  denominator  increases  without  limit,  the  value  of 
the  fraction  approaches  the  limit  0. 

It  is  customary  to  express  this  principle  as  follows  T 

«=0. 


78  ALGEBRA 

128.  No  literal  meaning  can  be  attached  to  such  results  as 

a  a      A 

-  =  oo ,  or  —  =  0  ; 

for  there  can  be  no  such  thing  as  division  unless  the  divisor  is 
a  finite  number. 

If  such  forms  occur  in  mathematical  investigations,  they 
must  be  interpreted  as  indicated  in  §§  126  and  127.  (Coin- 
pare  §  86.) 

THE  PROBLEM  OF  THE  COURIERS 

129.  The  following  discussion  will  further  illustrate  the 
form  -,  besides  furnishing  an  interpretation  of  the  form  -• 

The  Problem  of  the  Couriers. 

Two  couriers,  A  and  B,  are  travelling  along  the  same  road  in 
the  same  direction,  RB',  at  the  rates  of  m  and  n  miles  an  hour, 
respectively.  If  at  any  time,  say  12  o'clock,  A  is  at  P,  and  B 
is  a  miles  beyond  him  at  Q,  after  how  many  hours,  and  how 
many  miles  beyond  P,  are  they  together  ? 

B  P  Q  Rl 

I I I l 

Let  A  and  R  meet  x  hours  after  12  o'clock,  and  y  miles  beyond  P. 

They  will  then  meet  y  —  a  miles  beyond  Q. 

Since  A  travels  mx  miles,  and  B  nx  miles,  in  x  hours,  we  have 

f         y  —  m#, 

iy  —  a  =  nx. 

Solving  these  equations,  we  obtain 


We  will  now  discuss  these  results  under  different  hypotheses. 

1.   m>n. 

In  this  case,  the  values  of  x  and  y  are  positive. 

This  means  that  the  couriers  meet  at  some  time  after  12,  at  some  point 
to  the  right  of  P. 


FRACTIONS  79 

This  agrees  with  the  hypothesis  made  ;  for  if  m  is  greater  than  n.  A  is 

travelling  faster  than  B  ;  and  he  must  overtake  him  at  some  point  beyond 

their  positions  at  12  o'clock. 

2.  m<n. 

In  this  case,  the  values  of  x  and  y  are  negative. 

This  means  that  the  couriers  met  at  some  time  before  12,  at  some  point 
to  the  left  of  P. 

This  agrees  with  the  hypothesis  made;  for  if  m  is  less  than  n,  A  is 
travelling  more  slowly  than  B ;  and  they  must  have  been  together  before 
12  o'clock,  and  before  they  could  have  advanced  as  far  as  P. 

3.  a  =  0,  and  m  >  n  or  m  <  n. 

In  this  case,  x  —  0  and  y  =  0. 

This  means  that  the  travellers  are  together  at  12  o'clock,  at  the  point  P. 

This  agrees  with  the  hypothesis  made  ;  for  if  a  =  0,  and  m  and  n  are 
unequal,  the  couriers  are  together  at  12  o'clock,  and  are  travelling  at 
unequal  rates  ;  and  they  could  not  have  been  together  before  12,  and  will 
not  be  together  afterwards. 

4.  m  =  n,  and  a  not  equal  to  0. 

In   this  case,   the  values  of  x  and  y  take   the   forms  -  and  —.  re- 

*•     i  0  0 

spectively. 

If  m  —  n  approaches  the  limit  0,  the  values  of  x  and  y  increase  without 
limit  (§  126)  ;  hence,  if  m  =  n,  no  fixed  values  can  be  assigned  to  x  and  y, 
and  the  problem  is  impossible. 

In  this  case,  the  result  in  the  form  -  indicates  that  the  given  problem  is 
impossible. 

This  agrees  with  the  hypothesis  made  ;  for  if  m  =  n,  and  a  is  not  zero, 
the  couriers  are  a  miles  apart  at  12  o'clock,  and  are  travelling  at  the  same 
rate  ;  and  they  never  could  have  been,  and  never  will  be  together. 

5.    m  =  rc,  and  a  =  0. 

In  this  case,  the  values  of  x  and  y  take  the  form  -• 

If  a  =  0,  and  m  =  n,  the  couriers  are  together  at  12  o'clock,  and  travel- 
ling at  the  same  rate. 

Hence,  they  always  have  been,  and  always  will  be,  together. 

In  this  case,  the  number  of  solutions  is  indefinitely  great  ;  for  any 
value  of  x  whatever,  together  with  the  corresponding  value  of  y,  will 
satisfy  the  given  conditions. 

In  this  case,  the  result  in  the  form  -  indicates  that  the  number  of  solu- 
tions is  indefinitely  great. 

Such  form  is  called  Indeterminate. 


80  ALGEBRA 

130.  In  §  129,  we  found  that  the  form  -  indicated  an  ex- 
pression which*  could  have  any  value  ivhatever;  but  this  is  not 
always  the  case. 

Consider,  for  example,  the  fraction  x  ~~  a  • 

x2  —  ax 

If  x  —  a,  the  fraction  takes  the  form  -• 

0 

Now,  x2-a?  _  (x  +  a)(x-a)  __  x+a  . 

x2  — ax  ic(x  —  a)  x 

which  last  expression  is  equal  to  the  given  fraction  provided  x  does  not 
equal  a. 

The  fraction  x  +  a  approaches  the  limit  a  "*"  a,  or  2,  when  x  approaches 
the  limit  a.  x  a 

This  limit  we  call  the  value  of  the  given  fraction  ivhen  x  =  a. 

Then,  the  value  of  the  given  fraction  when  x  =  a  is  2. 

In  any  similar  case,  we  cancel  the  factor  which  equals  0  for  the  given 
value  of  ac,  and  find  the  limit  approached  by  the  result  when  x  approaches 
the  given  value  as  a  limit. 

EXERCISE   28 

Find  the  values  of  the  following : 

2  ax  —  4  a2     i  o  x2  — 16        ,  A 

2  or3 -5  a;2     .  A  4a;2 -4a; -3      , 

2.  when  a?  =0.        4. — when.r=i|. 

4:X2  +  3x  •       6af'-17a;  +  12 

s.    — — 2—  when  x  =  —  2. 

D         a**-8a!*-f-16 

6.   —         -~J—-    when  x  =  2. 

ar*  -  7  a;  -f  G 

131.   Other  Indeterminate  Forms. 

Expressions  taking  the  forms  ||-,  0  x  00  ,  or  qo  —  00,  for  oer- 
tain  values  of  the  letters  involved,  are  also  iiulct erniiiiate. 


FRACTIONS  81 

i.  Find  the  value  of  (ar3  +  8)  (l  +  -i— \  when  x  =  -  2. 

This  expression  takes  the  form  0  x  oo,  when  x  =—  2  (§  126). 

Now,  (x3  +  8)  ( 1  +  -i-^  =  x3  +  8  +  ^-ii? 
V        £  +  2/  x  +  2 

,   =  ic3  +  8  +  x2-2x  +  4  =  x3+x2  — 2x+12. 

The  latter  expression  approaches  the  limit  —  8+4  +  4  +  12,  or  12, 
when  x  approaches  the  limit  —  2. 

This  limit  we  call  the  value  of  the  expression  when  x  =—  2  ;  then,  the 
value  of  the  expression  when  x  =  —  2,  is  12. 

In  any  similar  case,  we  simplify  as  much  as  possible  before  finding  the 
limit. 

1  2x 

2.  Find  the  value  of --  when  x  =  1. 

1  —  x     1  —  x2 

The  expression  takes  the  form  oo  —  oo,  when  x  =  1  (§  126). 

Now  1  2x    :=l+x-2x=  1-s  =     1      r 

1  -  x      1-x2  1-x2         1  -  x2      1  +  x 

The  latter  expression  approaches  the  limit  J  when  x  approaches  the 
limit  1. 

Then,  the  value  of  the  expression  when  x  =  1,  is  \. 

132.   Another  example  in  which  the  result  is  indeterminate 

is  the  following : 

1  4-  2x 
Ex.  Find  the  limit  approached  by  the  fraction ^—  when 

x  is  indefinitely  increased. 

Both  numerator  and  denominator  increase  indefinitely  in  absolute  value 

when  x  is  indefinitely  increased. 

1  +  2 
1  +  2x     x 
Dividing  each  term  of  the  fraction  by  x,  = =—  =  = 

A  —  OX        L 

O 

X 

The  latter  expression  approaches  the  limit  r  (§  127),  or  —  -,  when 

x  is  indefinitely  increased. 

In  any  similar  case,  we  divide  both  numerator  and  denominator  of  the 
fraction  by  the  highest  power  of  x. 


82  ALGEBRA 


EXERCISE   29 

Find  the  limits  approached  by  the  following  when  x  is  in- 
definitely increased : 

'     4  +  5a?-3s8_  liii.  a^-2a?-4 

Find  the  values  of  the  following : 
1  12       *  o 

5.    (2a;2-5#-3)f2  +  -i-Vvhena  =  3. 
V        X  —  3J 


RATIO   AND   PROPORTION 
RAtlO 

133.  The  Ratio  of  one  number  a  to  another  number  b  is  the 

quotient  of  a  divided  by  b. 

Thus,  the  ratio  of  a  to  b  is  -;  it  is  also  expressed  a :  b. 

b 

The  ratios  here  spoken  of  are  but  fractions  under  another 
name,  and  have  all  the  properties  of  fractions. 

In  the  ratio  a:  b,  a  is  called  the  first  term,  or  antecedent,  and 

b  the  second  term,  or  consequent. 

If  a  and  b  are  positive  numbers,  and  a  >  b,  -  is  called  a 

b 

ratio  of  greater  inequality ;  if  a  <  5,  it  is  called  a  ra/10  of  less 
inequality. 

134.  A  ratio  of  greater  inequality  is  decreased,  and 
one  of  less  inequality  is  increased,  by  adding  the  same 
positive  number  to  each  of  its  terms. 

Let  a  and  b  be  positive  numbers,  a  being  >  b,  and  x  a  positive  number. 
Since  a  >  &,  ax  >  bx.  (§  50) 

Adding  ab  to  both  members  (§  50), 

ab  4-  ax>ab  +  bx,  or  a(&  -f  x)>b(a  +x). 


RATIO   AND    PROPORTION  83 


Dividing  both  members  by  b(b  -f  x),  we  have 


j»j±|.  _  (|M) 


In  like  manner,  if  a  <  6,       -  < 'L±*. 
6      6  +  x 


PROPORTION 

135.  A  Proportion  is  an  equation  whose  members  are  equal 
ratios. 

Thus,  if  a  :  b  and  c  :  d  are  equal  ratios, 

a:b  =  c  :d,  or  -  =e  - , 

is  a  proportion.     The  latter  form  is  preferable. 

136.  In  the  proportion  a:  b  =  c :  d,  a  is  called  the  first  term, 
b  the  second,  c  the  third,  and  d  the  fourth. 

The  first  and  third  terms  of  a  proportion  are  called  the  ante- 
cedents, and  the  second  and  fourth  terms  the  consequents. 

The  first  and  fourth  terms  are  called  the  extremes,  and  the 
second  and  third  terms  the  means. 

137.  If  the  means  of  a  proportion  are  equal,  either  mean  is 
called  the  Mean  Proportional  between  the  first  and  last  terms, 
and  the  last  term  is  called  the  Third  Proportional  to  the  first 
and  second  terms. 

Thus,  in  the  proportion  a:b  =  b  :  c,  b  is  the  mean  proportional 
between  a  and  c,  and  c  is  the  third  proportional  to  a  and  b. 

The  Fourth  Proportional  to  three  numbers  is  the  fourth  term 
of  a  proportion  whose  first  three  terms  are  the  three  numbers 
taken  in  their  order. 

Thus,  in  the  proportion  a:b  =  c:d,  d  is  the  fourth  proportional  to 
a,  b,  and  c. 

138.  A  Continued  Proportion  is  a  series  of  equal  ratios,  in 
which  each  consequent  is  the  same  as  the  next  antecedent;  as, 

a  :  b  =  b  \  c  =  c :  d  =  d :  e. 


«_ 
6  = 

~  d 

ad : 

=  bc. 

84  ALGEBRA 

PROPERTIES   OF   PROPORTIONS 

139.  In  any  proportion,  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Let  the  proportion  be 

Clearing  of  fractions, 

140.  From  the  equation  ad  =  bc  (§  139),  we  obtain 

be    ,      ad  ad        -.   7      be 

a  =  — ,  o  =  — ,  c  =  — ,  and  a  =  —  • 
deb  a 

That  is,  in  any  proportion,  either  extreme  equals  the 
product  of  the  means  divided  by  the  other  extreme ;  and 
either  mean  equals  the  product  of  the  extremes  divided 
by  the  other  mean. 

141 .  (Converse  of  §  139.)  If  the  product  of  two  numbers 
be  equal  to  the  product  of  two  others,  one  pair  may  be 
made  the  extremes,  and  the  other  pair  the  means,  of  a 
proportion. 

Let  ad  =  be. 

Dividing  by  bd,  ^  =  H ,  or  ■  ff  =  5 . 

°    J  bd     bd*       b     d 

In  like  manner,  we  may  prove  that 

a_b 
c      d"1 

c-  =  f,etc. 
d     b 

142.  In  any  proportion,  the  terms  are  in  proportion  by 
Alternation;  that  is,  the  means  may  be  interchanged. 

Let  the  proportion  be 

Then,  by  §  139, 

Then,  by  §  141, 

In  like  manner  it  may  be  proved  that  the  extremes  can  be  interchanged. 


a  _ 
b~ 

c 
'  d 

ad  = 

■be. 

«  _ 

_bt 
~  d 

a  _ 
b~ 

c 
d1 

ad  = 

-.be. 

b 

d 

RATIO   AND   PROPORTION  85 

143.  In  any  proportion,  the  terms  are  in  proportion  by 
Inversion  ;  that  is,  the  second  term  is  to  the  first  as  the 
fourth  tef  m  is  to  the  third. 

Let  the  proportion  be 
Then,  by  §  139, 

Whence,  by  §  141, 

a      c 

It  follows  from  §  143  that,  in  any  proportion,  the  means  can  be  written 
as  the  extremes,  and  the  extremes  as  the  means. 

144.  The  mean  proportional  between  two  numbers  is 
equal  to  the  square  root  of  their  product. 

Let  the  proportion  be  -  =  -  • 

b     c 

Then,  by  §  139,  b2  =  ac,  or  b  =  Vac. 

145.  In  any  proportion,  the  terms  are  in  proportion 
by  Composition ;  that  is,  the  sum  of  the  first  two  terms 
is  to  the  first  term  as  the  sum  of  the  last  two  terms  is 
to  the  third  term. 

Let  the  proportion  be  -  =  -  • 

b     d 

Then,  ad  =  be. 

Adding  each  member  of  the  equation  to  ac, 

ac  +  ad  =  ac  +  be,  or  a(c  +  d)  —  c(a  +  b). 

By  §141,  a±b  =  e_±dt 

a  c 

ttt  i  a  +  b     c  -f  d 

We  may  also  prove  — r —  =  — ;— 

b  d 

146.  In  like  mariner  we  may  also  prove  that  the  terms  of 
any  proportion  are  in  proportion  by  Division ;  that  is,  the  dif- 
ference between  the  first  two  terms  is  to  the  first  terin  as  the 
difference  between  the  last  two  terms  is  to  the  third  term. 

The  proof  is  left  to  the  student. 


86  ALGEBRA 

147.  In  any  proportion,  the  terms  are  in  proportion  by 
Composition  and  Division  ;  that  is,  the  sum  of  the  first 
two  terms  is  to  their  difference  as  the  sum  of  the  last 
two  terms  is  to  their  difference. 

The  proof  is  left  to  the  student.  Hint.  —  Divide  the  result  of  §  145  by 
that  of  §  146. 

148.  In  any  proportion,  if  the  first  two  terms  be  multi- 
plied by  any  number,  as  also  the  last  two,  the  resulting 
numbers  will  be  in  proportion. 

Let  the  proportion  be  -  =  - ;  then,   2£  =  *£. 

b      d  mb      nd 

(Either  m  or  n  may  be  unity  ;  that  is,  the  terms  of  either  ratio  may  be 
multiplied  without  multiplying  the  terms  of  the  other.) 

149.  In  any  proportion,  if  the  first  and  third  terms  be 
multiplied  by  any  number,  as  also  the  second  and  fourth 
terms,  the  resulting  numbers  will  be  in  proportion. 

Let  the  proportion  be  -  =  - ;  then,  —  =  —  • 

b      d  nb      nd 

(Either  m  or  n  may  be  unity.) 

1 50.  In  any  number  of  proportions,  the  products  of  the 
corresponding  terms  are  in  proportion. 

Let  the  proportions  be  -  —  -,  and   -  3=  .*!■" 

b      d  f      h 

Multiplying,  *  x  ?  =  ^  X  #r  OT  2«  =  ££. 

b     f     d      h  bf      dh 

In  like  manner,  the  theorem  may  be  proved  for  any  number  of 
proportions. 

151.  In  any  proportion,  like  powers  or  like  roots  of  the 
terms  are  in  proportion. 

Let  the  proportion  be  -  =  -  ;  then,   —  =  —  • 

b      d  bn      d» 

nt  n,~ 

In  like  manner,  = • 

Vb       Vd 


RATIO   AND    PROPORTION  87 

152.  In  a  series  of  equal  ratios,  any  antecedent  is  to  its 
consequent  as  the  sum  of  all  the  antecedents  is  to  the 
sum  of  all  the  consequents. 


Let 

a:  b  =  c.d  =  e:f. 

Then,  by  §  139, 

ad  —  6c, 

and 

af  =  be. 

Also, 

ab  =  ba. 

Adding, 

a(b+d+f)=b(a  +  c  +  e). 

Whence, 

a  :b  =  a  +  c  +  e  \b  +  d  +  f. 

(§  141) 

In  like  manner,  the  theorem  may  be  proved  for  any  number  of  equal 

ratios. 

153.  If  three  numbers  are  in  continued  proportion, 
the  first  is  to  the  third  as  the  square  of  the  first  is 
to  the  square  of  the  second. 

Let  the  proportion  be    a  :  b  =  b  :  c  ;  or  -  =  - . 

b      c 

Then,  ?x^?X«,or?  =  «!. 

b      c      b      b         c      b2 

154.  If  four  numbers  are  in  continued  proportion, 
the  first  is  to  the  fourth  as  the  cube  of  the  first  is 
to  the   cube  of  the  second. 

Let  the  proportion  be     a  :b  =  b  :c  =  c  :d  ;  or  -  =*  -  =  -. 

bed 

Then,  2x*x£=^xfx$|OT  ?»£ 

b     e      d      b      b      b         d     63 

Similarly,  it  may  be  shown  that  if  n  numbers  are  in  continued  propor- 
tion, the  first  antecedent  is  to  the  last  consequent  as  the  »th  power  of  the 
first  antecedent  is  to  the  nth  power  of  its  consequent. 

155.  Examples. 

i.  If  x  :  y  =  (x  +  zf :  (y  +  z)2,  prove  z  the  mean  proportional 
between  x  and  y. 

From  the  given  proportion,  by  §  139, 

y(x  +  z)2  =  x(y  +  z)\ 
Or,  x2y  4-  2  xyz  +  yz2  =  xy2  4-  2  xyz  +  xz2. 

Transposing,  x2y  —  xy2  =  xz2  —  yz2. 

Dividing  by  x  —  ?/,  xy  —  z2. 

Therefore,  z  is  the  mean  proportional  between  x  and  y  (§  144). 


88  ALGEBRA 

The  theorem  of  §  147  saves  work  in  the  solution  of  a  certain 
class  of  fractional  equations. 

2.    Solve  the  equation  2  X  ±  |  =  ?A^. 
^  2z-3      26  +  a 

Regarding  this  as  a  proportion,  we  have  by  composition  and  division, 


4  x  _   4b 
6  ~  -'la 

2x 

26. 
j 
a 

whence,  x 

86 

a 

3- 

Pro 

ve  that  if  - : 
b 

c 

then 

a2-b2: 

a2- 

■3ab-- 

=  c2- 

-d2: 

c2- 

Scd. 

Let 

«  _ 
6~ 

-  =  x,  whence 
d 

i,  a  = 

-  bx  ;  then, 

c2 

-1 

a2  -  b2 
a2- Sab     I 

b2x/ 

W  - 

!-62 
-3  6% 

X2 

-  1 

d2 

_  c2 
c2- 

-tf2 

X2- 

-3x 

c2 

3c 

-  o  Cd 

d2 

d 

Then, 

a2-62 

:a2 

-3a6 

=  c2 

-d2 

:  c2  — 

3  6"d. 

EXERCISE   30 

i.  Find  the  mean  proportional  between  .0289  and  1.69. 

2.  Find  the  mean  proportional  between  lT7^r  and  12||. 

3.  Find  the  third  proportional  to  if  and  1|. 

4.  Find  the  fourth  proportional  to  9g\,  16^,  and  T9g. 

5.  Find  the  fourth  proportional  to  m,  n,  and  r. 

6.  Write  in  the  form  of  a  proportion :    x2  —  2  x  -*  15  =  a2. 

Solve,  using  composition  and  division: 
4#-f-5_a?-f5 
4  #  —  5      a?  —  3 

x  —  a      b  —  c  2x  —  3      5  #  —  9 

ai  +  a     3  (m  +  l)2-(w-lV 

12.    If  -  =  - ,  show  that  a:c  =  b2:  c2. 
6      c 


8. 


RATIO   AND    PROPORTION  .        89 

•  i3.  fl^ry^cVtg-A^iaArt; »^^^y^»^^a^a»^-»r< 

14.  Find  two  numbers  in  the  ratio  of  2:.'{,  sucli  that  the 
sum  of  their  squares  shall  be  208. 

15.  Find  two  numbers  in  the  ratio  of  3 : 1,  such  that  the  dif- 
ference of  their  squares  is  200. 

16.  Two  numbers  are  in  the  ratio  of  5  :  7.  If  6  be  added  to 
each,  they  will  be  in  the  ratio  of  7  :  9.     Find  the  numbers. 

17.  Two  numbers  are  in  the  ratio  of  2  :  5.  If  4  be  added  to 
each  number,  the  resulting  ratio  will  be  twice  the  ratio  had  4 
been  subtracted  from  each  number.     Find  the  numbers. 

18.  The  difference  between  two  numbers  is  6,  and  the  dif- 
ference between  their  squares  is  60.  What  is  the  ratio  of  their 
sum  to  their  difference  ? 

19.  In  similar  figures  in  geometry,  homologous  sides  are  pro- 
portional. If  a  pole  30  feet  high  casts  a  shadow  42  feet  long, 
how  high  must  a  pole  be  to  cast  a  shadow  35  feet  long  ? 

20.  A  ladder  40  feet  long  leans  against  the  side  of  a  build- 
ing, with  its  foot  12  feet  from  the  building.  A  second  ladder, 
40^  feet  long,  makes  the  same  angle  with  the  building  as  the 
first  ladder.  How  far  is  the  foot  of  the  second  ladder  from 
the  building  ? 

21.  In  the  triangle  ABC,  MN  is 
drawn  parallel  to  BC  and  divides  the 
other   two    sides    proportionally.     If 

AM =12,  4M=2    and  5(7=48,  how 

'  AN     3' 

long  is  AC?     (M  is  the  middle  point  of  AB.)     What  is  the 
ratio  of  AN  to  MN? 

22.  The  areas  of  any  two  similar  figures  are  to  each  other 
as  the  squares  of  their  homologous  .parts.  If  a  regularTiexagon 
has  a  side  equal  to  6  and  an  area  of  54  V3,  what  is  the  area  of 
a  regular  hexagon  whose  side  is  2  ? 


90  ALGEBRA 

23.  The  area  of  a  circle  is  6^  times  that  of  another  circle. 
If  the  radius  of  the  first  circle  is  5,  what  is  the  radius  of  the 
second  circle  ? 

24.  If  the  altitude  of  a  triangle  is  twice  that  of  a  similar 
triangle,  how  do  their  areas  compare  ? 

25.  The  volume  of  a  rectangular  solid  is  equal  to  the  product 
of  its  three  dimensions,  x,  y,  and  z.  If  xyz  —  v  and  x  :  y  :  z 
=5  a  :  b  :  c,  filid  to,  y,  and  2  in  terms  of  a,  b,  c,  and  v. 

26.  Find  three  numbers  in  continued  proportion  whose  sum 
is  63,  the  second  being  4  times  the  first. 

27.  Given  the  proportion  -  =  -  =  -,  where  d  =  81  and  -  =  -  . 
_.    ,       .         ,  bed  bo 
Find  a,  0,  and  c. 

28.  If  2  a  —  36:4a-56  =  26— 3  c:46  —  5  c,  prove  6  is  the 
mean  proportional  to  a  and  c. 

29.  If3a  +  56:4a-76  =  3c  +  5d:4c-7d,  prove  ?==^ 

6     a 

30.  Find  two  numbers  in  the  ratio  of  a  to  6,  such  that  if 
3  increased  by  -  they  will  be  in  the  rat: 

ix  +  7     8x  +  ±     12x  +  l_Jx-l         Solvefora;. 


each  be  increased  by  -  they  will  be  in  the  ratio  of  e  to  /. 


31 


15  45  9(5 »  +  2) 


fl±6  +  q-2fta(2a^&)g  +  3a6i     Solve  for  ^ 
a?  ^  +  a  a2  —  a2 

33.  A  man  borrows  a  certain  sum,  paying  interest  at  the 
rate  of  5%.  After  repaying  $180,  his  interest  rate  on  the 
balance  is  reduced  to  4J%,  and  his  annual  interest  is  now  less 
by  $  10.80.     Find  the  sum  borrowed. 

34.  The  digits  of  a  certain  number  are  three  consecutive 
numbers,  of  which  the  middle  digit  is  the  greatest,  and  the 
first  digit  the  least.  If  the  number  be  divided  by  the  sum  of 
its  digits,  the  quotient  is  ^p.     Find  the  number. 


VARIATION  91 

35.  A  certain  number  of  apples  were  divided  between  three 
boys.  The  first  received  one-half  the  entire  Dumber,  with  one 
apple  additional,  the  second  received  one-third  the  remainder, 
with  one  apple  additional,  and  the  third  received  the  remain- 
der, 7.     How  many  apples  were  there  ? 

36.  A  freight  train  runs  6  miles  an  hour  less  than  a  pas- 
senger train.  It  runs  80  miles  in  the  same  time  that  the 
passenger  train  runs  112  miles.     Find  the  rate  of  each  train. 

37.  A  and  B  each  fire  40  times  at  a  target ;  A's  hits  are  one- 
half  as  numerous  as  B's  misses,  and  A's  misses  exceed  by  15 
the  number  of  B's  hits.  How  many  times  does  each  hit  the 
target  ? 

t  38.  A  freight  train  travels  from  A  to  B  at  the  rate  of  12 
miles  an  hour.  After  it  has  been  gone  31  hours,  an  express 
train  leaves  A  for  B,  travelling  at  the  rate  of  45  miles  an  hour, 
and  reaches  B  1  hour  and  5  minutes  ahead  of  the  freight. 
Find  the  distance  from  A  to  B,  and  the  time  taken  by  the 
express  train. 

39.  A  tank  has  three  taps.  By  the  first  it  can  be  filled  in 
3  hours  10  minutes,  by  the  second  it  can  be  filled  in  4  hours 
45  minutes,  and  by  the  third  it  can  be  emptied  in  3  hours 
48  minutes.  How  many  hours  will  it  take  to  fill  it  if  all  the 
taps  are  open? 

40.  A  man  invested  a  certain  sum  at  3|%,  and  \\  this  sum 
at  4^%  ;  after  paying  an  income  tax  of  5%,  his  net  annual 
income  is  $  195.70.     How  much  did  he  invest  in  each  way  ? 

VARIATION 

1 56.  One  variable  number  (§  124)  is  said  to  vary  directly  as 
another  when  the  ratio  of  any  two  values  of  the  first  equals 
the  ratio  of  the  corresponding  values  of  the  second. 

It  is  usual  to  omit  the  word  "  directly  •'  and  simply  say  that  one  nnin 
ber  varies  as  another. 


92  ALGEBRA 

Thus,  if  a  workman  received  a  fixed  number  of  dollars  per 
diem,  the  number  of  dollars  received  in  m  days  -will  be  to  the 
number  received  in  n  days  as  til  is  to  n. 

Then,  the  ratio  of  any  two  numbers  of  dollars  received 
equals  the  ratio  of  the  corresponding  numbers  of  days  worked. 

Hence,  the  number  of  dollars  which  the  workman  receives 
varies  as  the  number  of  days  during  which  he  works. 

157.  The  symbol  co  is  read  "varies  as"  ;  thus,  ace  b  is  read 
"  a  varies  as  6." 

158.  One  variable  number  is  said  to  vary  inversely  as 
another  when  the  first  varies  directly  as  the  reciprocal  of  the 
second. 

Thus,  the  number  of  hours  in  which  a  railway  train  will 
traverse  a  fixed  route  varies  inversely  as  the  speed ;  if  the 
speed  be  doubled,  the  train  will  traverse  its  route  in  one-half 
the  number  of  hours. 

159.  One  variable  number  is  said  to  vary  as  two  others 
jointly  when  it  varies  directly  as  their  product. 

Thus,  the  number  of  dollars  received  by  a  workman  in  a 
certain  number  of  days  varies  jointly  as  the  number  which  he 
receives  in  one  day,  and  the  number  of  days  during  which  he 
works. 

160.  One  variable  number  is  said  to  vary  directly  as  a  sec- 
ond and  inversely  as  a  third,  when  it  varies  jointly  as  the 
second  and  the  reciprocal  of  the  third. 

Thus,  the  attraction  of  a  body  varies  directly  as  the  amount 
of  matter,  and  inversely  as  the  square  of  the  distance. 

161 .  Ifxccy,  then  x  equals  y  multiplied  by  a  constant  number. 

Let  x'  and  y'  denote  a  fixed  pair  of  corresponding  values  of  x  and  y, 
and  x  and  y  any  other  pair. 

By  the  definition  of  §  156,  -  =  - ;  or,  x  -  -y. 

y    y'  yr 

x' 
Denoting  the  constant  ratio  —  by  m,  we  have 

y' 

x  =  my. 


VARIATION  93 

162.  It  follows  from  §§  158,  159,  160,  and  161  that : 

1.  If  x  varies  inversely  as  y,  x  =  —  • 

y 

2.  If  x  varies  jointly  as  y  and  z,  x  =  myz. 

3.  If  x  varies  directly  as  y  and  inversely  as  z,  a?  =  ^  • 

z 

1 63.  Ifxccy,  and  y  oc  z,  then  xccz. 

By  §  161,  if  x  cc  y,  x  =  my.  (1) 

And  iiyazz,  y  —  nz. 

Substituting  in  (1),  x  —  mnz. 

Whence,  by  §  161,  x*z. 

164.  Ifxccy  when  z  is  constant,  and  xccz  when  y  is  constant, 
then  xcc  yz  when  both  y  and  z  vary. 

Let  y'  and  z'  be  the  values  of  y  and  z,  respectively,  when  x  has  the 
value  x'. 

Let  y  be  changed  from  yf  to  y",  z  remaining  constantly  equal  to  z\  " 
and  let  x  be  changed  in  consequence  from  x'  to  X. 

Then,  by  §156,  ^=T/'  W 

Now,  let  z  be  changed  from  z'  to  z",  y  remaining  constantly  equal  to 
y",  and  let  x  be  changed  in  consequence  from  Xto  x  . 

Then,  —  =  — .  (2) 

x"      z"  K  J 

Multiplying  (1)  by  (2),  ±  =  &L>  (3) 

xn     y"z" 

Now  if  both  changes  are  made,  that  is,  y  from  y'  to  y"  and  z  from  z'  to 
z",  x  is  changed  from  xf  to  x",  and  yz  is  changed  from  y'z'  to  u"z". 

Then  by  (3),  the  ratio  of  any  two  values  of  x  equals  the  ratio  of  the 
corresponding  values  of  yz  ;  and,  by  §  156,  xccyz. 

The  following  is  an  illustration  of  the  above  theorem  : 

It  is  known,  by  Geometry,  that  the  area  of  a  triangle  vaties  as  the 
base  when  the  altitude  is  constant,  and  as  the  altitude  when  the  ba.se  is 
constant;  hence,  when  both  base  and  altitude  vary,  the  area  varies  as 
their  product. 


94  ALGEBRA 

165.    Problems. 

Problems  in  variation  are  readily  solved  by  converting  the 
variation  into  an  equation  by  aid  of  §§  161  or  162. 

i.  If  x  varies  inversely  as  y,  and  equals  9  when  y  =  S,  find 
the  value  of  x  when  y  =  18. 

If  x  varies  inversely  as  y,  x  =—  (§  162). 

y 

Putting  x  =  9  and  y  =  8,  9  =  — ,  or  m  =  72. 
8 

Then,  x  =  — ;  and,  if  y  =  18,  x  =  —  =  4. 
V  18 

Since  variation  is  simply  another  way  of  stating  a  proportion,  the  prob- 
lems in  variation  may  be  solved  readily  by  means  of  proportion. 

E.g.     In  the  above  problem  ^ 

xcc  -, 

y 

x.=  ™. 
y 

This  equation  is  true  for  any  assigned  values  of  the  variables. 

Then,         ■  xx=-,  (1) 

2/i 

x2=™-  (2) 

,      2/2 

Dividing  (1)  by  (2)  Q  =  V*  (3) 

x2     2/i 

which  is  in  the  form  of  inverse  proportion.     Substituting  the  given  values 
of  x  and  y  in  (3) ,  we  have  q      ..  8 

x2      8  ' 

9  .  8 

whence  x2  =  ■ =  4. 

18 

2.    Given  that  the  area  of   a  triangle  varies  jointly  as  its 

base  and  altitude,  what  will  be  the  base  of  a  triangle  whose 

altitude  is  12,  equivalent  to  the  sum  of  two  triangles  whose 

bases  are  10  and  6,  and  altitudes  3  and  9,  respectively  ? 

Let  J5,  //,  and  A  denote  the  base,  altitude,  and  area,  respectively,  of 
any  triangle,  and  B'  the  base  of  the  required  triangle. 

Since  A  varies  jointly  as  />  and  //,  A  =  mlUI  (§  162). 

Therefore,  the  area  of  the  first  triangle  is  m  x  10  x  3,  or  30  ???,,  and  the 
area  of  the  second  is  m  x  6  x  9,  or  54  m. 


VARIATroX  95 

Then,  the  area  of  the  required  triangle  is  30  m  -f  54  m,  or  84  m. 
But,  the  area  of  the  required  triangle  is  also  m  x  B'  x  12. 

Therefore,  12  mB'  =  84  m,  or  B'  =  7. 

Or  using  proportion  and  letting  A\  —  area  of  first  triangle,  A2  =  area  of 
second,  As  =  area  of  third. 

A3  =  Ai  +  A2 

Ai  =  mB1H1.  (l) 

A2  =  mB2H2.  (2) 

As  =  mB3H3.  (3) 

Adding  (1)  and  (2) 

Ax  +  A2  =  m^BiHi  +  B2H2).  (4) 

Dividing  (4)  by  (3) 

Ai+A2  =  m^BxHi  +  B2II2) 
As  m(BsH3) 

or,  1=BlHl  +  B*H3m  (5) 

Substituting  the  given  values  of  B  and  if  in  (5)  we  have 
1  =  10  .  3  +  6  ■  9 
12  £3       ' 
whence,  B3  =  7. 

EXERCISE  31 

i.   If  xvzy,  and  #  =  3  when  ?/ =  12,  what  is  the  value  of  x 
when  y  =  2S? 

2.  If  y&x2,  and  2/  =  4  when  x  =  l,  what  is  the  value  of  y 
in  terms  of  x2  ? 

3.  If  2/  varies  inversely  as  x,  and  y  =  4  when  #  ==  —  3,  wrhat 
is  the  value  of  y  when  x  =  2  ? 

4.  If  &  varies  directly  as  y  and  inversely  as  2,  and  x  =  £ 
when  #  =  -§  and  3  =  f ,  what  is  the  value  of  x  when  y  =  £  and 

5.  If  a;  varies  jointly  as  y  and  z  and  #  =  —  20  when  y  =  2 
and  3  =  8,  what  is  the  value  of  #  when  ?/  =  —  £  and  z<==  16  ? 

6.  If  (3  x  +  4)  oc  (2  ?/  —  5)  when  #  =  —  1  and  y  =  4,  wThat  is 
the  value  of  x  when  ?/  =  19  ? 


96  ALGEBRA 

7.  If  x2  varies  inversely  as  y8,  when  x  =  4  and  ?/  =  2f  what 
is  the  value  of  2/  when  a?  =  ?/f  ? 

8.  If  x  equals  the  sum  of  two  numbers,  one  of  which  varies 
directly  as  y  and  the  other  inversely  as  z2,  and  x  =  47  when 
y  =  —  16  and  3  =  2,  and  a;  =  2  when  y  =  —  2  and  z  =  1,  find  the 
value  of  x  when  y  =  3  and  z  =  \* 

9.  The  area  of  a  triangle  varies  jointly  as  its  base  and 
altitude.  If  the  area  of  a  triangle  whose  base  is  6  and  whose 
altitude  is  9  is  27,  what  is  the  base  of  a  triangle  whose  area  is 
44  and  whose  altitude  is  11  ? 

10.  The  distance  through  which  a  body  falls  from  rest 
varies  as  the  square  of  the  time  during  which  it  falls.  If  a 
body  falls  900  feet  in  7.5  seconds,  how  many  feet  will  it  fall 
in  16  seconds  ? 

11.  The  illumination  from  a  source  of  light  varies  inversely 
as  the  square  of  the  distance  from  the  source.  How  far  must 
an  object  20  feet  from  the  light  be  moved  in  order  that  it  may 
receive  twice  as  much  light  ? 

12.  A  circular  plate  of  lead,  17  inches  in  diameter,  is  melted 
and  formed  into  three  circular  plates  of  the  same  thickness. 
If  the  diameters  of  two  of  the  plates  are  8  and  9  inches 
respectively,  find  the  diameter  of  the  other;  it  being  given 
that  the  area  of  a  circle  varies  as  the  square  of  its  diameter. 

13.  A  cow  tied  to  a  stake  by  a  rope  24  yards  long  will  graze 
over  the  area  within  her  reach  in  three  days.  She  breaks  her 
rope  and,  in  repairing  it,  it  is  shortened  1^  feet.  In  how  many 
days  will  she  graze  over  the  new  area  ? 

14.  A  pump  supplying  the  water  for  a  building  has  a  10-ineh 
stroke  and  a  cylinder  4  inches  in  diameter.  It  is  not  possible 
to  increase  the  number  of  strokes  of  the  pump,  nor  to  increase 
the  length  of  the  cylinder.  By  how  much  must  the  diameter 
be  increased  if  50%  is  added  to  the  capacity  of  the  pump'/ 
(The  volumes  of  cylinders  vary  as  the  product  of  the  base  and 
altitude.) 


INVOLUTION    AND    EVOLUTION  97 

VI.     INVOLUTION   AND   EVOLUTION 

166.  We  have  already  given  (Chapter  III)  the  involution 
and  evolution  of  monomials.  We  will  now  consider  involution 
and  evolution  of  polynomials. 

167.  Square  of  a  polynomial.     By  actual  multiplication 

(a  +  b  +  c)2  =  a2  +  b2  +  c2  +  2  ab  +  2  ac  +  2  be. 
In  like  manner 
(a  +  b  +  c  +  df 

=  a2  +  &Vf-  c2  +  d2  +  2  a&  +  2  ac  +  2  ad  +  2  6c  +  2  6cZ  +  2  cd, 
and  so  on  for  the  square  of  any  polynomial. 
The  law  observed  may  be  stated  as  follows : 

The  square  of  a  polynomial  is  equal  to  the  sum  of 
the  squares  of  its  terms,  together  with  twice  the  prod- 
uct of  each  term  by  each  of  the  following  terms. 

Ex.     Expand  (2  x2  -  3  x  -  5)2. 

The  squares  of  the  terms  are  4  sc4,  9  x2,  and  25. 

Twice  the  product  of  the  first  term  by  each  of  the  following  terms  gives 
the  results  —  12  x:3  and  -  20  x2. 

Twice  the  product  of  the  second  term  by  the  following  term  gives  the 
result  30  x. 

Then,     (2  x2  -  3  x  -  5)2  =  4  x4  +  9  x2  +  25  -  12  x*  -  20  x2  +  30  x 
=  4  x4  -  12  cc3  -  11  x2  +  30  x  +  25. 

168.  Cube  of  a  binomial.     By  actual  multiplication 

(a  -f  bf  =  d3  +  3  a2b  +  3  a&2  +  Jft 

That  is,  the  cube  of  the  sum  of  two  numbers  is  equal  to 
the  cube  of  the  first,  plus  three  times  the  square  of  the 
first  times  the  second,  plus  three  times  the  first  times 
the  square  of  the  second,  plus  the  cube  of  the  second. 

In  like  manner,  the  cube  of  the  difference  of  two 
numbers  is  equal  to  the  cube  of  the  first,  minus  three 
times  the  square  of  the  first  times  the  second,  plus  three 
times  the  first  times  the  square  of  the  second,  minus 
the  cube  of  the  second. 

The  cube  of  a  trinomial  may  be  found  by  the  above  method, 
if  two  of  its  terms  be  enclosed  in  parenthesis,  and  regarded  as 
a  single  term. 


98  ALGEBRA 

169.  Square  Root  of  any  Polynomial  Perfect  Square. 

By  §  167,  (a  +  b  +  cf  =  a2  +  2  ab  +  V1  +  2  ac  +  2  be  +  c2 

=  a2  +  (2a  +  &)6  +  (2a  +  2&  +  c)e.  (1) 

Then,  if  the  square  of  a  trinomial  be  arranged  in  order  of 
powers  of  some  letter : 

I.  The  square  root  of  the  first  term  gives  the  first  term  of 
the  root,  a. 

II.  If  from  (1)  we  subtract  a2,  we  have 

(2a  +  6)H(2«  +  2Hc)c       #  (2) 

The  first  term  of  this,  when  expanded,  is  2  ab ;  if  this  be 
divided  by  twice  the  first  term  of  the  root,  2  a,  we  have  the 
next  term  of  the  root,  b. 

III.  If  from  (2)  we  subtract  (2a  +  b)b,  we  have 

(2a  +  2&  +  c)c.  (3) 

The  first  term  of  this,  when  expanded,  is  2  ac;  if  this  be 
divided  by  twice  the  first  term  of  the  root,  2  a,  we  have  the 
last  term  of  the  root,  c. 

IV.  If  from  (3)  we  subtract  (2a  +  2&  +  c)c,  there  is  no 
remainder. 

Similar  considerations  hold  with  respect  to  the  square  of  a 
polynomial  of  any  number  of  terms. 

170.  The  principles  of  §  169  may  be  used  to  find  the  square 
root  of  a  polynomial  perfect  square  of  any  number  of  terms. 

Let  it  be  required  to  find  the  square  root  of 

4  x4  +  12  x3  -  7  x2  -  24  x  +  16. 

4  x4  +  12  xs  -    7  x2  -  24  x  +  16    1  2  x2  +  3  x  -  4 
a2  =  4  a* 


2  a  +  &  =  4  x2  +  3  .x 

a  a 


12  x8-    7  a2  -  24  X  +  16,  1st  Rem. 


2a+2&  +  c  =  4.r2  +  6x  —  4 
-4 


-  16  x2  -  24  x  4-  16,  2d  Rem. 
_  16x*  — 24x  +  16 


2  x2  +  3  x  —  4  is  called  the  square  root  and  2  a  the  first  trial  divisor. 
2  a  +  b  is  the  first  complete  divisor. 


INVOLUTION   AND  EVOLUTION  99 

We  then  have  the  following  rule  for  extracting  the  square 
root  of  a  polynomial  perfect  square : 

Arrange  the  expression  according  to  the  powers  of 
some  letter. 

Extract  the  square  root  of  the  first  term,  write  the 
result  as  the  first  term  of  the  root,  and  subtract  its 
square  from  the  given  expression,  arranging  the  re- 
mainder in  the  same  order  of  powers  as  the  given  ex- 
pression. 

Divide  the  first  term  of  the  remainder  by  twice  the 
first  term  of  the  root,  and  add  the  quotient  to  the  part 
of  the  root  already  found,  and  also  to  the  trial  divisor. 

Multiply  the  complete  divisor  by  the  term  of  the  root 
last  obtained,  and  subtract  the  product  from  the  re- 
mainder. 

If  other  terms  remain,  proceed  as  before,  doubling  the 
part  of  the  root  already  found  for  the  next  trial  divisor. 

171.    Cube  Root  of  any  Polynomial  Perfect  Cube. 

By  §168,  (a+&  +  c)3=[(a  +  6)+c]3 
=  (a  -f  bf  +  3(a  +  b) 2c  +  3(«  +  b)c2  -f.c3 
=  a3  +  3  a2b  +  3  ab2  +  V  +  3(a  +  b)2c  +  3(a  +  by  +  c3 
=  a3  +  (3  a2+  3  ab  +  b2)lj  +  [3(«  +  bf+  3(a  +  b)c  +  c^c.  (1) 

Then,  if  the  cube  of  a  trinomial  be  arranged  in  order  of 
powers  of  some  letter  : 

I.  The  cube  root  of  the  first  term  gives  the  first  term  of  the 
cube  root,  a. 

II.  If  from  (1)  we  subtract  a3,  we  have 

(3a2  +  3ab  +  b2)b  +  [3(a  +  b)2  +  3(a  +  b)c  +  c2]c.       (2) 

The  first  term  of  this,  when  expanded,  is  3a26;  if  this  be 
divided  by  three  times  the  square  of  the  first  term  of  the  root, 
3a2,  we  have  the  next  term  of  the  root,  b. 

III.  If  from  (2)  we  subtract  (3  a2  +  3  ab  4-  &*)&,  we  have 

I3(a  +  by  +  3(a  +  b)c  +  c2]c.  (3) 


100  ALGEBRA 

The  first  term  of  this,  when  expanded,  is  3a2c;  if  this  be 
divided  by  three  times  the  square  of  the  first  term  of  the  root, 
3  <r,  we  have  the  last  term  of  the  root,  c.  ■ 

IV.  If  from  (3)  we  subtract  [3(a  +  &)2  +  3  (a  +  ^'"  +  «2>, 
there  is  no  remainder. 

Similar  considerations  hold  with  respect  to  the  cube  of  poly- 
nomials of  any  number  of  terms. 

172.  The  principles  of  §  171  may  be  used  to  find  the  cnbe 
root  of  a  polynomial  perfect  cube  of  any  number  of  terms. 

Let  it  be  required  to  find  the  cnbe  root  of 

aj6  +  6a5  +  3iB4-28aj8-9a?2  +  54aj-27. 


x6H-6^+3x4-28x3-9x2+o4x-27 

=  x6 


3  «2+3  ab  +  b2  =  3  x*+6  x3+4  x2 

2x 


6x5  +  3x*-28x3-9x2+54x-27 

0xr>  +  12x*4-'8x3 


S(a  +  b)2  =  Sx^  +  12xii+12x2 
3(a  +  6)c  +  c'2=_  -  9x2-18x+9 

8s*  +  12a;*+  3x2-18x+9 


-  9;e4_;30x3-9x2  +  54x-27 

-  9x4-36\x3-9x2+54x-27 


The  first  term  of  the  root  is  the  cube  root  5f  x6,  or  x2. 

Subtracting  the  cube"  of  x2,  or  x6,  from  the  given  expression,  the  first 
remainder  is  6  x5  +  3  x4  —  28  x3  —  9  x2  +  54  x  —  27. 

Dividing  the  first  term  of  this  by  three  times  the  square  of  the  first 
term  of  the  root,  3x4,  we  have  the  next  term  of  the  root,  2x  (§  171,  II). 

Now,  3  ab  +  b2  equals  3  x  x2  x  2  x  +  (2  x)2,  or  6  x3  +  4  x2. 

Adding  this  to  3x4,  multiplying  the  result  by  2x,  and  subtracting  the 
product,  6x5  +  12  x4  +  8x3,  from  the  first  remainder,  gives  the  second 
remainder,  -  9x4  -  30  x3-  9x2  -f  54  x-  27  (§  171,  III). 

Dividing  the  first  term  of  this  by  three  times  the  square  of  the  first 
term  of  the  root,  3x2,  we  have  the  last  term  of  the  root,  —  3. 

Now,  8(a  +  b)2  equals  3(x2  +  2  x)2,  or  3  x4  +  12x3  +  12  x2;  3(«  +  b)c 
equals  3(x2  +  2x)(  —  3),  or  —  9x2—  18  x  ;  and  tfi  =  9. 

Adding  these  results,  we  have  3x4  -f  12  x3  +  3x2  —  18  x  +  9. 

Subtracting  from  the  second  remainder  the  product  of  this  by  —  8,  <n- 
—  9  x*  —  86  .''"'  —  0 x-  -f  54  x  —  27,  there  is  no  remainder  ;  then,  x2  +  2  %  —  8 
is  the  required  root  (J  171,  W). 

The  expressions  :>»./'t  and  :>>  .'"*  -f 12  Xs  +  12  x2  are  called  trial  divisor*, 
and  the  expressions  3  x4  +  0  x3  -f  4  x2  and  3  x4  +  12  x8  +  3  x2  —  18  x  -f  9 
complete  divisors. 


INVOLUTION    AM)    EVOLUTION  101 

We  then  have  the  following  rule  for  finding  the  cube  root  of 
a  polynomial  perfect  cube  : 

Arrange  the  expression  according  to  the  powers  of 
some  letter. 

Extract  the  cube  root  of  the  first  term,  write  the  result 
as  the  first  term  of  the  root,  and  subtract  its  cube  from 
the  given  expression ;  arranging  the  remainder  in  the 
same  order  of  powers  as  the  given  expression. 

Divide  the  first  term  of  the  remainder  by  three  times 
the  square  of  the  first  term  of  the  root,  and  write  the 
result  as  the  next  term  of  the  root. 

Add  to  the  trial  divisor  three  times  the  product  of  the 
term  of  the  root  last  obtained  by  the  part  of  the  root 
previously  found,  and  the  square  of  the  term  of  the  root 
last  obtained. 

Multiply  the  complete  divisor  by  the  term  of  the 
root  last  obtained,  and  subtract  the  product  from  the 
remainder. 

If  other  terms  remain,  proceed  as  before,  taking  three 
times  the  square  of  the  part  of  the  root  already  found  for 
the  next  trial  divisor. 

EXERCISE   32 
Find  the  square  roots  of  the  following : 

i.    4  a4  +  12  a*b  -  7  a2b2  -  24  db*  +  16  b\ 

2.  49m4-5m2-42m3  +  l  +  6m. 

3.  9a2-24a&-36«c  +  1662  +  48&c  +  36c2. 

5.  x6  +  5xi  +  Uxs-6x5  +  l-4x-2x2. 

6.  4m4-25mi-12m*  +  16-24,y,l. 

7.  64  c2 -80  c -23  + 9c"2  +  30  c"1. 

8.  4  x  +  9  y~A  +  4  xhj~2  +  24  $#"*  -  1 0  %  V1- 

9.  6  yz~2  +  ±x-2  +  y2-4  x-hj  - 12  x~lz~2  +  9  z~\ 


102  ALGEBRA 

10.    4  a3  +  29  a  -  4  J  +  21  a2  -  20  a  *  +  4  - 18  a* 

Find  the  cube  roots  of  the  following : 

ii.   343 %*  -  441  x>y  4  189  xy2 -  27  y\ 

12.   a6 -9  0^4  21  a4 4  9 ar* - 42 a?2- 36 ar- 8. 

13..  18  a4  -  13  a3  +  1  +  8  aG  4  9  a2  -  3  a  - 12  a5. 

14.    54  m5  4  44  ra3  +  1  +  27  mfi  4  63  m4  4  6  m  4  21  m2. 

5  3  3      27 

16.  64  ah-6  -  240  a/>"4c  +  300  ah~2c2  -  125  c3. 

17.  8  .s3  4  36  s2  4  18  «  -  81  -  27  s"1  4  81  s~2  -  27  s~3: 

1 8.  21  a*  -  54  a*  +  27  ai  +  63  a  -  44  a*  4 1  -  6  a*. 

19.  a?"3  -  3  x~hf  4  3  ar1?/  -  z3  -  3  ar2z  -.y*  4-  6  arty*«  -  3#s 

+  3  x-  V  _  3  y  ^ 

20.  a  4  6  aV1  4  12  ah~2  4-  8  6~3  4  3  a  V2  4  12  aVV"2 

+  12  6"2c-2  4  3  a  V4  4-  6  Ir^r*  4  c"6. 

Find  the  fourth  roots  of  the  following : 

21.  81  a10  -  36  a*kx~*  4  6  a5*-1*0  - $  a*ar5  4  gV  *"*• 

22.  a?8-  12 x-7  4  50a;6  -72 x5  -21  a*4  4- 72 a8  +  50 a2 4  12 a +  1. 

Find  the  sixth  roots  of  the  following : 

23.  64  m12  -  192  m10  4-  240  m8  -  160  m«  4-  60  m4  -  1 2  ms  4  1. 

24.  a3  -  3  a*6*  4  V  <**  -  f  a*(t  4  1  j  «?;6  -  A  «^¥  +  *l*  &9- 

173.    Square  Root  of  any  Integral  Perfect  Square. 
The  square  root  of  an  integral  perfect  square  may  be  found 
in  the  same  way  as  the  square  root  of  a  polynomial. 


INVOLUTION   AND    EVOLUTION  103 

We  have  the  following  rule  for  finding  the  square  root  of  an 
integral  perfect  square : 

Separate  the  number  into  periods  of  two  digits  each, 
beginning  with  the  units'  place. 

Find  the  greatest  square  in  the  left-hand  period,  and 
write  its  square  root  as  the  first  digit  of  the  root ;  sub- 
tract the  square  of  the  first  root  digit  from  the  left-hand 
period,  and  to  the  result  annex  the  next  period. 

Divide  this  remainder,  omitting  the  last  digit,  by  twice 
the  part  of  the  root  already  found,  and  annex  the  quo- 
tient to  the  root,  and  also  to  the  trial  divisor. 

Multiply  tne  complete  divisor  by  the  root  digit  last 
obtained,  and  subtract  the  product  from  the  remainder. 

If  other  periods  remain,  proceed  as  before,  doubling 
the  part  of  the  root  already  found  for  the  next  trial 
divisor. 

Note  1 :  It  sometimes  happens  that,  on  multiplying  a  complete  divisor 
by  the  digit  of  the  root  last  obtained,  the  product  is  greater  than  the 
remainder. 

In  such  a  case,  the  digit  of  the  root  last  obtained  is  too  great,  and  one 
less  must  be  substituted  for  it. 

Note  2 :  If  any  root  digit  is  0,  annex  0  to  the  trial  divisor,  and  annex 
to  the  remainder  the  next  period. 

Ex.   Required  the  square  root  of  15376. 


1 

a2  =  1 
2  a  +  b  =  200  +  20 
b  =            20 

'53'76 
00  00 
53  76 
44  00 

100  +  20  +  4 

=  a  +  b  +  c 

=  (2a  +  b)b 

2a  +  26  +  c  =  200+40  +  4 
4 

9  76 
9  76 

=  (2a  +  2b  +  c)c 

Pointing  the  number  in  accordance  with  the  rule  of  §  173,  we  find  that 
there  are  three  digits  in  its  square  root. 

Let  a  represent  the  hundreds'  digit  of  the  root,  with  two  ciphers 
annexed  ;  b  the  tens'  digit,  with  one  cipher  annexed ;  and  c  the  units' 
digit. 

Then,  a  must  be  the  greatest  multiple  of  100  whose  square  is  less  than 
15376  ;  this  we  find  to  be  100. 

Subtracting  a2,  or  10000,  from  the  given  number,  the  result  is  5376. 


104  ALGEBRA 

Dividing  the  remainder  by  2  a,  or  200,  we  have  the  quotient  26+  ;  which 
suggests  that  b  equals  20. 

Adding  this  to  2  a,  or  200,  and  multiplying  the  result  by  b,  or  20,  we 
have  4400  ;  which,  subtracted  from  5376,  leaves  97(1. 

Since  this  remainder  equals  (2  a  +  2  b  +  c)c,  we  can  get  c  approxi- 
mately by  dividing  it  by  2  a  +  2  &,  or  200  +  40. 

Dividing  976  by  240,  we  have  the  quotient  4+  ;  which  suggests  that 
c  equals  4. 

Adding  this  to  240,  multiplying  the  result  by  4,  and  subtracting  the 
product,  976,  there  is  no  remainder. 

Then  124  is  the  square  root. 

Omitting  the  ciphers  for  the  sake  of  brevity,  and  condensing  the  opera- 
tion, we  may  arrange  the  work  of  the  example  as  follows: 

1'53'76[124 
1 


22 


53 
44 


2441976 
976 


CUBE  ROOT   OP   AN   ARITHMETICAL  NUMBER 

174.  The  cube  root  of  1000  is  10 ;  of  1000000  is  100,  etc. 

Hence,  the  cube  root  of  a  number  between  1  and  1000  is  be- 
tween 1  and  10 ;  the  cube  root  of  a  number  between  1000  and 
1000000  is  between  10  and  100 ;  etc. 

That  is,  the  integral  part  of  the  cube  root  of  an  integer  of 
one,  two,  or  three  digits  contains  one  digit;  of  an  integer  of 
four,  five,  or  six  digits  contains  two  digits ;  and  so  on. 

Hence,  if  a  point  be  placed  over  every  third  digit  of  an 
integer,  beginning  at  the  units'  place,  the  number  of 
points  shows  the  number  of  digits  in  the  integral  part 
of  its  cube  root. 

175.  Cube  Root  of  any  Integral  Perfect  Cube. 

The  cube  root  of  an  integral  perfect  cube  may  be  found  in 
the  same  way  as  the  cube  root  of  a  polynomial. 


INVOLUTION    AND    EVOLUTION  105 


Required  the  cube  root  of  124871G8. 

12487108 1  200  +  30  +  2 
q8  =  8000000 1  =  a  +   b  +  c 


3  a2  =  120000 

Sab  =    18000 

b1  =       900 


138900 
30 


4487168 


4167000 


3(a  +  b)2=    158700 

3(a  +  b)c  =       1380 

c2  =  4 


160084 

2 


320168 


320168 


Pointing  the  number  in  accordance  with  the  rule  of  §  174,  we  find 
that  there  are  three  digits  in  the  cube  root. 

Let  a  represent  the  hundreds'  digit  of  the  root,  with  two  ciphers 
annexed ;  b  the  tens'  digit,  with  one  cipher  annexed ;  and  c  the  units' 
digit. 

Then,  a  must  be  the  greatest  multiple  of  100  whose  cube  is  less  than 
12487168  ;  this  we  find  to  be  200. 

Subtracting  a3,  or  8000000,  from  the  given  number,  the  result  is 
4487168. 

Dividing  this  by  3  a2,  or  120000,  we  have  the  quotient  37+  ;  which  sug- 
gests that  b  equals  30. 

Adding  to  the  divisor  120000,  3  ab,  or  18000,  and  62,  or  900,  we  have 
138900. 

Multiplying  this  by  &,  or  30,  and  subtracting  the  product  4167000  from 
4487168,  we  have  320168. 

Since  this  remainder  equals  [3 (a  +  &)2  +  3(a  +  b)c  +  c2]c  (§  171,  III), 
we  can  get  c  approximately  by  dividing  it  by  3(a  +  ft)2,  or  168700. 

Dividing  320168  by  158700,  the  quotient  is  2+  ;  which  suggests  that  c 
equals  2. 

Adding  to  the  divisor  158700,  3 (a  +  6)c,  or  1380,  and  c2,  or  4,  we  have 
160084  ;  multiplying  this  by  2,  and  subtracting  the  product,  320168,  there 
is  no  remainder. 

Then,  200  +  30  +  2,  or  232,  is  the  required  cube  root. 

176.  Omitting  the  ciphers  for  the  sake  of  brevity^  and  con- 
densing the  process,  the  work  of  the  example  of  §  175  will 
stand  as  follows ; 


106  ALGEBRA 


1200 

180 

9 

1389 

4487 
4167 

158700 

1380 

4 

160084 

320168 
320168 

The  numbers  120000  and  158700  are  called  trial  divisors,  and  the  num- 
bers 138900  and  160084  are  called  complete  divisors. 

We  then  have  the  following  rule  for  finding  the  cube  root  of 
an  integral  perfect  cube : 

Separate  the  number  into  periods  by  pointing  every 
third  digit,  beginning  with  the  units'  place. 

Find  the  greatest  cube  in  the  left-hand  period,  and 
write  its  cube  root  as  the  first  digit  of  the  root ;  subtract 
the  cube  of  the  first  root  digit  from  the  left-hand  period, 
and  to  the  result  annex  the  next  period. 

Divide  this  remainder  by  three  times  the  square  of 
the  part  of  the  root  already  found,  with  two  ciphers 
annexed,  and  write  the  quotient  as  the  next  digit  of  the 
root. 

Add  to  the  trial  divisor  three  times  the  product  of  the 
last  root  digit  by  the  part  of  the  root  previously  found, 
with  one  cipher  annexed,  and  the  square  of  the  last  root 
digit. 

Multiply  the  complete  divisor  by  the  digit  of  the  root 
last  obtained,  and  subtract  the  product  from  the  re- 
mainder. -* 

If  other  periods  remain,  proceed  as  before,  taking 
three  times  the  square  of  the  part  of  the  root  already 
found,  with  two  ciphers  annexed,  for  the  next  trial 
divisor. 

Note  1 :  Note  1,  §  173,  applies  with  equal  force  to  the  above  rule. 
Note  2 :  If  any  root-figure  is  0,  annex  two  ciphers  to  the  trial  divisor, 
and  annex  to  the  remainder  the  next  period. 


INVOLUTION   AND   EVOLUTION  107 

177.  In  the  example  of  §  175,  the  first  complete  divisor  is 

3a2+3ab  +  b2.  (1) 

The  next  trial  divisor  is  3  (a  -f  b)2,  or  3  a2  -f  6  ab  +  3  b2. 

This  may  be  obtained  from  (1)  by  adding  to*  it  its  second 
term,  and  double  its  third  term. 

That  is,  if  the  first  number  and  the  double  of  the  sec- 
ond number  required  to  complete  any  trial  divisor  be 
added  to  the  complete  divisor,  the  result,  with  two 
ciphers  annexed,  will  give  the  next  trial  divisor. 

This  rule  saves  much  labor  in  forming  the  trial  divisors. 

Ex.     Find  the  cube  root  of  157464. 

157464  |_54_ 
125 


7500 

600 

16 


8116 


32464 


32464 


EXERCISE   33 

Find  the  square  roots  of  the  following : 
i.  5294601.  3.  .00098596.  5.  .0037319881. 

2.  68.7241.  4.  567762.25. 

Find  the  cube  roots  of  the  following : 

6.  658503.  9.  .000070444997. 

7.  1953125.  10.  .000001601613. 

8.  748.613312. 

Find  the  first  four  figures  of  the  square  roots  of : 

11.   3.  12.   f  13.   if-  14.   f  15-   iM 

Find  the  first  four  figures  of  the  cube  roots  of : 

16.    5.  17.    16.  18.    J.  19-     -~-  20-   *V 


108  ALGEBRA 

OTHER  POWERS 

178.  A  Series  is  a  succession  of  terms. 

A  Finite  Series  is  one  having  a  limited  number  of  terms. 
An  Infinite.  Series   is  one  having   an  unlimited  number   of 
terms. 

179.  In  §§  103  and  168  we  gave  rules  for  finding  the  square 
or  cube  of  any  binomial. 

The  Binomial  Theorem  is  a  formula  by  means  of  which  any 
power  of  a  binomial  may  be  expanded  into  a  series. 

180.  Proof  of  the  Binomial  Theorem  for  a  Positive  Integral 
Exponent. 

The  following  are  obtained  by  actual  multiplication : 

(a  +  x)2  =  a2  +  2  ax  +  x2 ; 

(a  +  x)s  =  a?+3a2x  +  3ax2  +  xi',    . 

(a  +  x)4  =  a4  +  4  a3x  +  6  cPx2  -f  4  ax3  +  x4 ;  etc. 

In  these  results,  we  observe  the  following  laws : 

1.  The  number  of  terms  is  greater  by  1  than  the  exponent 
of  the  binomial. 

2.  The  exponent  of  a  in  the  first  term  is  the  same  as  the 
exponent  of  the  binomial,  and  decreases  by  1  in  each  succeed- 
ing term. 

3.  The  exponent  of  x  in  the  second  term  is  1,  and  increases 
by  1  in  each  succeeding  term. 

4.  The  coefficient  of  the  first  term  is  1,  and  the  coefficient  of 
the  second  term  is  the  exponent  of  the  binomial. 

5.  If  the  coefficient  of  any  term  be  multiplied  by  the  expo- 
nent of  a  in  that  term,  and  the  result  divided  by  the  exponent 
of  x  in  the  term  increased  by  1,  the  quotient  will  be  the 
coefficient  of  the  next  following  term. 

181.  If  the  laws  of  §  180  be  assumed  to  hold  for  the  expan- 
sion of  (a-f-  x)n,  where  n  is  any  positive  integer,  the  exponent 
of  a  in  the  first  term  is  n,  in  the  second  term  n  —  1,  in  the 
third  term  n  —  2,  in  the  fourth  term  n  —  3,  etc. 


INVOLUTION   AND   EVOLUTION  109 

The  exponent  of  x  in  the  second  term  is  1,  in  the  third  term 
2,  in  the  fourth  term  3,  etc. 

The  coefficient  of  the  first  term  is  1*;  of  the  second  term  n. 

Multiplying  the  coefficient  of  the  second  term,  n,  by  n  —  1, 
the  exponent  of  a  in  that  term,  and  dividing  the  result  by 
the  exponent  of  x  in  the  term  increased  by  1,  or  2,  we  have 

Vl\u~   )  as  the  coefficient  of  the  third  term  ;  and  so  on. 
1.2  ' 

Then,  (a  +  x)n  =  an  +  nan~lx  +  n^"^1)^-2x2 

n(n-l)(n-2)     _^,  ■  (1) 

1.2-3  V  ' 

Multiplying  both  members  of  (1)  by  a  +  a,  we  have 
(«  +  x)^1  =  «»+!  +  nanx  -4-  w(n~"  X)  an~W  +  WCW~  1)C»-2)g>-2g8  +  ... 

1  •  2i  1  •  '-  •  o 

+    anx+  nan~1x2  +  n(n  ~  ^  an~2x*  +  — . 

1   •  ii 

Collecting  the  terms  which  contain  like  powers  of  a  and  x,  we  have 
(a  +  z)n+1  =  an+1  +  (n  +  l)anx  +  pO*-1)  +  nla*-1*2 

rn(n-l)(n-2)  .  n(n-l)"]  ^^  ■  ... 
L         1-2.3  1-2     J 

=  a"*1  +  (n  +  l)anx  +  np-=-^  +  lla""1*2 

1-2     L    3  J 

Then,  (a  4-  x)n+1  =  ^n+1  +  O  4-  l)anx  +  n  r^-Ha""1*2 

=  a**1  +  (n  +  l)a"x  +  (n  +  *)*  o"-^2 

(w+l)ii(n-l)        2z3  +  ....  (2) 

1.2-3  K 


110  ALGEBRA 

It  will  be  observed  that  this  result  is  in  accordance  with 
the  laws  of  §  180 ;  which  proves  that,  if  the  laws  hold  for  any 
power  of  a  -f  x  whose  exponent  is  a  positive  integer,  they  also 
hold  for  a  power  whose  exponent  is  greater  by  1. 

But  the  laws  have  been  shown  to  hold  for  (a  -f-  x)4,  and 
hence  they  also  hold  for  (a  +  x)5;  and  since  they  hold  for 
(a  -f-  x)5,  they  also  hold  for  (a  +  #)6 ;  and  so  on. 

Therefore,  the  laws  hold  when  the  exponent  is  any  positive 
integer,  and  equation  (1)  is  proved  for  every  positive  integral 
value  of  n. 

Equation  (1)  is  called  the  Binomial  Theorem. 

In  place  of  the  denominators  1  •  2 ;  1-2-3,  etc.,  it  is  usual  to  write 
[2,  [3,  etc. 

The  symbol  |_n,  read  "  factorial  w,"  signifies  the  product  of  the  natural 
numbers  from  1  to  n,  inclusive. 


The  method  of  proof  exemplified  in  §  181  is  known  as  Mathematical 
Induction. 

182.  Putting  a  —  1  in  equation  (1),  §  181,  we  have 

(i+a?)»=i+n^+^r1>g»+n(n-yn--2)^+...- 

\2_  [3 

183.  In  expanding  expressions  by  the  Binomial  Theorem, 
it  is  convenient  to  obtain  the  exponents  and  coefficients  of  the 
terms  by  aid  of  the  laws  of  §  180. 

i .   Expand  (a  +  x)5. 

The  exponent  of  a  in  the  first  term  is  6,  and  decreases  by  1  in  each 
succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by  1  in  each 
succeeding  term. 

The  coefficient  of  the  first  term  is  1  ;  of  the  second,  5. 

Multiplying  5,  the  coefficient  of  the  second  term,  by  4,  the  exponent  of 
a  in  that  term,  and  dividing  the  result  by  the  exponent  of  x  increased  by 
1,  or  2,  we  have  10  as  the  coefficient  of  the  third  term  ;  avid  so  on. 

Then,     (a  +  a)5  =  a5  +  5  a4x  +  10  a'6x  +10  aV  +  6  axv  +  x6. 


INVOLUTION    AND   EVOLUTION  111 

It  will  be  observed  that  the  coefficients  of  terms  equally  distant  from 
the  ends  of  the  expansion  are  equal ;  this  law  will  be  proved  in  §  185. 

Thus  the  coefficients  of  the  latter  half  of  an  expansion  may  be  written 
out  from  the  first  half. 

If  the  second  term  of  the  binomial  is  negative,  it  should  be 
enclosed,  negative  sign  and  all,  in  parentheses  before  applying 
the  laws. 

2.  Expand  (1  —  xf. 

=  16  4-6.1&.  (-x)  +  15.1*.  (-a-)2 +  20- 1".  (-*)• 

+  15  •  I2  .  (  -  X)4  +  6  .  1  •  (-  X)5  +  (-  *)• 

=  1  -  6  x  +  15  x2  -20  x3  +  15  x4  -  6  x5  +  x6. 

If  the  first  term  of  the  binomial  is  an  arithmetical  number,  it  is  con- 
venient to  write  the  exponents  at  first  without  reduction  ;  the  result 
should  afterwards  be  reduced  to  its  simplest  form. 

If  either  term  of  the  binomial  has  a  coefficient  or  exponent 
other  than  unity,  it  should  be  enclosed  in  parentheses  before 
applying  the  laws. 

3.  Expand  (3  m2  —  Vn)4. 

(3  m2-  Vny  =  [(3  m2)  +  (-  ?ii)]4 

=  (3  ra2)*  +  4  (3  m2)3(  -  ni)  +  6(3  m2)2(-  ni)2 

+  4  (3  m2)  ( -  niy  +  (-  nty 
=  81  m8  -  108  m*f»£  +  54  m*n$  -  12  m2n  +  nl 

A  trinomial  may  be  raised  to  any  power  by  the  Binomial 
Theorem,  if  two  of  its  terms  be  enclosed  in  parentheses,  and 
regarded  as  a  single  term ;  but  for  second  powers,  the  method 
of  §  167  is  shorter. 

4.  Expand  (x2-2x- 2)4. 

(X2  _  2  x  -  2)4  =  [(x2  _  2x)  +  (-  2)]* 

=  (x2  -  2  x)4  +  4  (x2  -  2  x)3(-  2)  +  6  (x2-  2  x)2*;-  2)* 

+  4(x2-2x)(-2)3  +  (-2)4 
=  x8  -  8  x?  +  24  x«  -  32  x6  +  16x* 

_  8  ( x6  -  6  x5  +  1 2  x4  -  8  Xs ) 
+  24(x4-4x3  +  4x2)  -32(x2-2x)+  16 
=  x8  -  8  x1  +  16  xe  +  16  x5  -  56x4-32x3  +  64 x2  +  64x  +  16. 


112  ALGEBRA 

EXERCISE   34 

Expand  the  following : 

2.    (x-yf.  \3n  J 

3.  (i-*p.  ««•  (2c-*-icr^. 

W        ' '  16.  (1  -  a2)'2. 

5.    (a2  -  ft)\  \  _    '   ,  _ 

18.  (a  +  6)16. 

7.    (3  m  —  4  n)4.  ..  ~ 


8.  (^-2g)6.  I9'  (2a"t+i~=fJ 

9.  (ar2  +  ^)5  20.  (a^-7/V)11. 
10.    (2a~*  +  ^)7.  21.  (a  +  ^-c)4. 

/^              \6  22.  (x2  -  2  a  -  3)4. 


12. 


23.    (m2  -  2  m  +  I)4- 
a2     &2\8  24.    (a2  +  x  +  l)5. 


&       ay  25.    (1  +  c  +  c2) 


,2\6 


184.  To  find  the  rth  or  general  term  in  the  expansion  of 
(a  +  x)n. 

The  following  laws  hold  for  any  term  in  the  expansion  of 
(a  +  x)n,  in  equation  (1),  §  181 : 

1.  The  exponent  of  x  is  less  by  1  than  the  number  of  the 
term. 

2.  The  exponent  of  a  is  n  minus  the  exponent  of  x% 

3.  The  last  factor  of  the  numerator  is  greater  by  1  than  the 
exponent  of  a. 

4.  The  last  factor  of  the  denominator  is  the  same  as  the 
exponent  of  x. 

Therefore  in  the  rth  term,  the  exponent  of  x  will  be  r  —  1. 
The  exponent  of  a  will  be  n  —  (r  —  1),  or  n  —  r  +  1. 
The  last  factor  of  the  nu  111  crater  will  be  n  —  r  +  2. 
The  last  factor  of  the  denominator  will  be  r  —  1. 


INVOLUTION    AND   EVOLUTION  113 

Hence,  the  rth  term 

n(n  -  1)  O  -  2)  .  • .  (n  -  r  +  2)    n_r+1  r-1         w  i 

In  finding  any  term  of  an  expansion,  it  is  convenient  to  obtain 
the  coefficient  and  exponents  of  the  terms  by  the  above  laws. 

Ex.   Find  the  8th  term  of  (3  a*  -  &-1)11. 

We  have,  (3  ah  -  6"1)11  =  [(3  ah)  +  (-  ft"1)]11.    ■  ■ 

In  this  case,  n  =  11,  r  =  8. 

The  exponent  of  (—  b'1)  is  8  —  1,  or  7. 

The  exponent  of  (3  a%)  is  11  —  7,  or  4. 

The  first  factor  of  the  numerator  is  11,  and  the  last  factor  4  +  1,  or  5. 

The  last  factor  of  the  denominator  is  7. 

Then,  the  8th  term  =  U  •  10  ■  9-8- 7 -6- 5        Jw_  &_n7 
1.2.3.4.5.6.7^        ;i  ; 

=  330(81  a2)  (-  6"7)  =  -  26730  a2&"7. 

If  the  second  term  of  the  binomial  is  negative,  it  should  be  enclosed, 
sign  and  all,  in  parentheses  before  applying  the  laws. 

If  either  term  of  the  binomial  has  a  coefficient  or  exponent  other  than 
unity,  it  should  be  enclosed  in  parentheses  before  applying  the  laws. 

Find  the  :  exercise  35 

x.    5th  term  of  (a  +  b)K  ^  q{         _  JV* 

2.  Tth  term  of  (x-y)10.  3  xj  ' 

3.  6th  term  of  (1  -  a)11.  8.    6th  term  of  f^  + -^ 


lr 


4.  4th  term  of  (a2  —  &3)8. 

5.  8th  term  of  (d  -  2  #J»        g.    5th  term  of  (    /«  _Ji)\ 

6.    10th  term  of  (m's  +  r> )  *  ,.  ,1  *  p  /     /-      «      ,N7 

v  2/  io.   4th  term  of  (xVy  —  -f  y-i) 7. 

185.  Multiplying  both  terms  of  the  coefficient,  in  (1),  §  184, 
by  the  product  of  the  natural  numbers  from  1  to  n  —  r  -f- 1,  in- 
clusive, the  coefficient  of  the  rth  term  becomes 

w(n-l)--(n  — r  +  2)-(tt-r  +  l)-..2-l  [n 

\r  —  1  x  1  •  2  ...  (n -  r  +  1)  "  [r  —  1  \n  —  r  +  l ' 


114  ALGEBRA 

Since  the  number  of  terms  in  the  expansion  is  n  -j-  1,  the  rth 
term  from  the  end  is  the  (n  —  r  -f  2)th  from  the  beginning. 

Then,  to  find  the  coefficient  of  the  rth  term  from  the  end,  we 
put  in  the  above  formula  n  —  r  -f  2  for  r. 

Then,  the  coefficient  of  the  rth  term  from  the  end  is 


or 


\n  —  r  +  2  —  l  \n  —  (n  —  r  +  2)+l'        \n  —  r  +  1  \r  —  1 

Hence,  in  the  expansion  of  (a  +  a?)w,  the  coefficients  of 
terms  equidistant  from  the  ends  of  the  expansion  are 
equal. 

186.  It  was  proved  in  §  181  that,  if  n  is  a  positive  integer, 

(a  +  x)n  =  an  +  nan~lx  +  n(n  —  1)  an-2z2 
v  J  1-2 

,  n  (n  —  l)(n  —  2)    _-  o  . 
+    v    la2    q +  '"' 

If  n  is  a  negative  integer,  or  a  positive  or  negative  fraction, 
the  series  in  the  second  member  is  infinite  (§  178)  ;  for  no  one 
of  the  expressions  n  —  1,  n  —  2,  etc.,  can  equal  zero ;  in  this 
case,  the  series  gives  the  value  of  (a  +  x)n,  provided  it  is 
convergent. 

As  a  rigorous  proof  of  the  Binomial  Theorem  for  Fractional  and  Nega- 
tive Exponents  is  too  difficult  for  pupils  at  this  stage  of  their  progress,  the 
author  has  thought  best  to  omit  it ;  any  one  desiring  a  rigorous  algebraic 
proof  of  the  theorem,  will  find  it  in  the  author's  Advanced  Course  in 
Algebra,  §  575. 

187.  Examples. 

In  expanding  expressions  by  the  Binomial  Theorem  when 
the  exponent  is  fractional  or  negative,  the  exponents  and 
coefficients  of  the  terms  may  be  found  by  the  laws  of  §  180, 
which  hold  for  all  values  of  the  exponent. 

I.    Expand  (a  +  xy  to  five  terms. 

The  exponent  of  a  in  the  first  term  is  §,  and  decreases  by  1  in  each 
succeeding  term. 


INVOLUTION    AND   EVOLUTION  115 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by  1  in  each 
succeeding  term. 

The  coefficient  of  the  first  term  is  1  ;  of  the  second  term,  J. 

Multiplying  §,  the  coefficient  of  the  second  term,  by  —  |,  the  exponent 
of  a  in  that  term,  and  dividing  the  product  by  the  exponent  of  x  increased 
by  1,  or  2,  we  have  —  I  as  the  coefficient  of  the  third  term  ;  and  so  on. 

Then,  (a  +  «)*-  a%  +  f  ef*as  -  \a~^xL  +  ,^a~*x8  -  ^gT^o4  +  •••. 

2.  Expand  (1  +  2  x~^)~2  to  five  terms. 

Enclosing  2  z*  in  parentheses,  we  have 

(1+2  x~V2  =  [1  +  (2  aT*)]"2 

=  l"2  -  2  •  1-3  .  (2  x~i)  +  3  •  I"4  •  (2  x~i)2 

-  4  •  I"5  •  (2  x"i)3  +  5  •  I"6  •  (2  x~i)*  -  ••• 
=  l-4x"U  12z-1-32x"£  +  80x-2  + 

By  writing  the  exponents  of  1,  in  expanding  [1  -f  (2x~^)]-2,  we  can 
make  use  of  the  fifth  law  of  §  180. 

3.  Expand  to  four  terms. 

^V1  -  3  x* 

Enclosing  a-1  and  —  3  x*  in  parentheses,  we  have 

-_JL_  = J =  [(«-i)  +  (-3xi)]-* 

Va-i-3x*      (a"1  -3  a*)* 

=  (a"1)-*  -  1  (a-1)"^-  3  xi)  +  {  0-T^(-  3  xi)2 
-H(a-i)-:¥(-3a*)«+... 

=  as  +  afxi  +  2  afet  +  -1/-  a'^x  +  ♦••• 

EXERCISE   36 

Expand  each,  of  the  following  to  five  terms : 

1.  (a  +  x)*.  4-    Va-b.  7.    (a*  +  2  &)* 

1 

2.  (1 +  *)-*.  5*   (a  +  a;)5'  8-    (rf-4^)"*. 

3.  (1  —  x)  *.  VI  —  x  x~*  +  3  y 


116  ALGEBRA 


I2.    !___.    14.    (m*-3rr*)-3. 


10.    (m~3  +  * 


n.    V[(a~2— 66-c)7].  \2/      «/  xSva41  J 

188.  The  formula  for  the  rth  term  of  (a  +  x)n  (§  184)  holds 
for  fractional  or  negative  values  of  n,  since  it  was  derived  from 
an  expansion  which  holds  for  all  values  of  the  exponent. 

Ex.   Find  the  7th  term  of  (a  -  3  x~^)~K 

Enclosing  —  3  aft  in  parentheses,  we  have 

(a  -  3  x"i)"i  =  [a  +  (-  3  jc~f)]"i. 

The  exponent  of  (—  3  #~t)  is  7  —  1,  or  6. 
The  exponent  of  a  is  —  ±  —  6,  or  —  & 

The  first  factor  of  the  numerator  is  —  J,  and  the  last  factor  —  ^  +  1, 
or  -  ¥• 

The  last  factor  of  the  denominator  is  6. 
Hence,  the  7th  term 


1.23.4.5.6 

3»    v    y   9 


_.     ,     _  EXERCISE   37 

Find  the : 


a-V(-3ari)« 


1.    6th  term  of  (a  -f  *)*.  6.   11th  term  of  V(m  4-  n)5- 

K4,   .  -  ,        7N-i  7.  7th  term  of  (a~2-2  &V2. 

2.-  5th  term  of  (a  —  0)  ».  '  v  } 

3.  7th  term  of  (1  +  x)~\  S'   8th  term  of  ,_   *     r    ' 
*                     v  ^  ;  (rf  +  y  2)4 

4.  8th  term  of  (1  -  &)*.  9.  10th  term  of  (ar5  +  y  *)'* 

5.  9th  term  of  (a  -  x)~3.  10.    6th  term  of  (a*  -  2  6"4)"^ 

11.    5th  term  of  (m+3n-8)* 
1 


12.   9th  term  of 


^[(a3  +  3&-fy] 


INVOLUTION   AND   EVOLUTION  117 

189.    Extraction  of  Roots. 

The  Binomial  Theorem  may  sometimes  be  used  to  find  the 
approximate  root  of  a  number  which  is  not  a  perfect  power  of 
the  same  degree  as  the  index  of  the  root. 

Ex.   Find  V25  approximately  to  five  places  of  decimals. 

The  nearest  perfect  cube  to  25  is  27. 

We  have  v"25  =  ^27-2  =  [(38)  +  (  -  2)]* 

=  (33)}  +  i(38)"*(-  2)-  i(33)~*(-  2)2 


=  3-- 


+  A(33)  T(-2)«-. 
2  4  40 

3  .  32     9  •  35      81  •  38 


Expressing  each  fraction  approximately  to  the  nearest  fifth  decimal 
place,  we  have 

^/25  =  3  -  .07407  -  .00183  -  .00008 =  2.92402. 

We  then  have  the  following  rule : 

Separate  the  given  number  into  two  parts,  the  first  of 
which  is  the  nearest  perfect  power  of  the  same  degree  as 
the  required  root,  and  expand  the  result  by  the  Binomial 
Theorem. 

If  the  ratio  of  the  second  term  of  the  binomial  to  the  first  is  a  small 
proper  fraction,  the  terms  of  the  expansion  diminish  rapidly  ;  but  if  this 
ratio  is  but  little  less  than  1,  it  requires  a  great  many  terms  to  insure  any 
degree  of  accuracy. 

EXERCISE   38 

Find  the  approximate  values  of  the  following  to  five  places 
of  decimals : 

i.    Vl7.      2.    V51.      3.    ^60.     4.    ^14.     5.    \/84.     6.    a/35. 

PROPERTIES   OF   QUADRATIC   SURDS 

190.  A  quadratic  surd  (§  70)  cannot  equal  the  sum  of 
a  rational  expression  and  a  quadratic  surd. 

For,  if  possible,  let  (a)?  =  b  -f  (c)*, 

1  1 

where  6  is  a  rational  expression,  and  (a)2  and  (c)1  quadratic  surds. 


118  ALGEBRA 

Squaring  both  members,       a  =  62  +  2  6(c)'2  -f  c, 
or,  2  6(c)*  =  a-  62  -  c. 

Whence,  (c)  z  =  a  ~  ft2  ~  c . 

26 

That  is,  a  quadratic  surd  equal  to  a  rational  expression. 

But  this  is  impossible  ;  whence,  (a)2  cannot  equal  6  +(c)2. 

191.  If  a  +(&)*  =  c  +  (cf )*,  where  a  and  c  are  rational  ex- 
pressions, and  (5)*  and  (<f)*  quadratic  surds,  then 

a  =  c,  and  (&)*  -  («!)*. 
If  a  does  not  equal  c,  let  a  =  c  +  £  ;  then,  #  is  rational. 
Substituting  this  value  in  the  given  equation, 

c  +  aj-f  (6)*  =  c  +  (d)*,  or  £+(&)*  =(d)*. 

But  this  is  impossible  by  §  190. 

i      ,      i 
Then,  a  =  c,  and  therefore  (6)2  =(d)'z. 

192.  If  (a  +  V6)^  as  Vx  +  Vy,  where  a,  b,  x,  and  y  are  ra- 
tional expressions,  then  {a  -  VS) 2  ==  Vas  -  Vy. 

Squaring  both  members  of  the  given  equation, 

a  +  V6  =  x  +  2V:*;?/  +  y, 
Whence,  by  §  191,  a  =  x  +  y, 

and  (6)1  s=  2(xy)* 

Subtracting,  a  —  (6)2  =  x  —  2(scy)*  +  y. 

Extracting  the  square  root  of  both  members, 

(a  -y/ly  -Vx-y/y. 

193.  Square  Root  of  a  Binomial  Surd. 

The  preceding  principles  may  be  used  to  find  the  square 
roots  of  certain  expressions  which  are  in  the  form  of  the  sum 
of  a  rational  expression  and  a  quadratic  surd. 


INVOLUTION   AND   EVOLUTION  119 

Ex.    Find  the  square  root  of  13  —  Vl60. 


Assume, 

Vl3  -  V160  =  Vx  -  Vy. 

(1) 

Then,  by  §  192, 

Vl3  +  VUK)  =  Vx  +  Vy. 

(2) 

Multiply  (1)  by  (2), 

V169  -  160  =  x  —  f. 

Or, 

x  -  y  =  3. 

(3) 

Squaring  (1) 

13-VlOO  =x-2Vx2/ 

+  2/. 

Whence,  by  §  191, 

x  +  y  =  13. 

W 

Adding  (3)  and  (4), 

2#=  16,  or  x  = 

:8. 

Subtracting  (8)  from 

(4),                2?/ =  10,  ory  = 

:5. 

Substitute  in  (1),  Vl3  -  V160  =  V8  -  V5  =  2 V2  -  V5. 

194.  Examples  like  that  of  §  193t  may  be  solved  by  inspec- 
tion, by  putting  the  given  expression  into  the  form  of  a  tri- 
nomial perfect  square  (§  103,, II),  as  follows: 

Reduce  the  surd  term  so  that  its  coefficient  may  be  2. 

Separate  the  rational  term  into  two  parts  whose  prod- 
uct shall  be  the  expression  under  the  radical  sign  of  the 
surd  term. 

-     Extract  the  square  root  of  each  part,  and  connect  the 
results  by  the  sign  of  the  surd  term. 

i.    Extract  the  square  root  of  8  +V48. 

We  have  V48  =  2VT2. 

We  then  separate  8  into  two  parts  whose  product  is  12. 

The  parts  are  6  and  2  ;  whence, 


V8  +  V48  =  V6  +  2V12  +  2  =  v'6  +  VS. 
2.   Extract  the  square  root  of  22  —  3  V32. 


We  have       3  V32  =\/9x8x4=  2V72. 

We  then  separate  22  into  two  parts  whose  product  is  72. 

The  parts  are  18  and  4  ;  whence, 


V22  -3V32  =Vl8-2V72  +  4  =  Vl8  -  v/4  =  3V/2-2. 


120  AUJKIJKA 

EXERCISE  39 

Find  the  square  roots  of  the  following : 

i.  5  +  2V6.  9.    2c-2(c2-d2)K 

2.  8  —  2V12.  io.    m  +  2Vmn- n2. 

3-  8a-2aVl5.  II#    a _  Va2 -  £2. 

4.  9  +  2(14)1  I2.   5»  +  a>V21. 

5.  7  +  4(3)*  13.   113-12(85)1 

6.  17-12V2.  14.   366  +  24V2I0. 

7.  2  + (3)4.  15.   540 -14  VII. 

8.  l  +  i  V3. 

195.    Solution  of  Equations  having  the  Unknown  Numbers  under 
Radical  Signs. 


1.    Solve  the  equation  V#2  —  5  —  x  =  —  1. 


Transposing  —  x,  Vx2  —  5  =  x  —  1. 

Squaring  both  members,  cc2  —  5  =  #2  —  2j+1. 

Transposing,  2  x  =  G  ;  whence,  g  =  3. 

(Substituting  3  for  x  in  the  given  first  member,  and  taking  the  positive 
value  of  the  square  root,  the  first  member  becomes 


V9"^5" -  3  =  2  -  3  =  -  1  ; 
which  shows  that  the  solution  x  =  3  is  correct.) 

We  then  have  the  following  rule : 

Transpose  the  terms  of  the  equation  so  that  a  surd  term 
may  stand  alone  in  one  member;  then  raise  both  mem- 
bers to  a  power  of  the  same  degree  as  the  surd. 

If  surd  terms  still  remain,  repeat  the  operation. 

The  equation  should  be  simplified  as  much  as  possible  before  perform 
ing  the  involution. 


INVOLUTION   AND   EVOLUTION  121 


2.    Solve  the  equation  V2  x  —  1  -+■  V2  x  +  6  =  7. 


Transposing  V2  x  —  1,     a/2  x  +  6  =  7  —  V2  x  —  1. 


Squaring,  2x4-0  =  49  -  14  V2  x  -  1  +  2  x  —  1. 


Transposing,  14  v  2  x  —  1  =  42,  or  V2  x  —  1  =  3. 

Squaring,  2  x  —  1  =  9  ;  whence,  x  =  5. 

3.    Solve  the  equation  Vcf  —  2  —  Va7= 


V#  —  2 


Clearing  of  fractions,      x  —  2  —  Vx2  —  2  x  =  1. 


Transposing,  —  Vx'2  —  2  x  =  3  —  x. 

Squaring,  x2  —  2  x  =  9  —  6  x  +  x2. 

Transposing,  4  x  =  9,  and  x  =  -  • . 

4 

0 

(If  we  put  x  =  - ,  the  given  equation  becomes 

If  we  take  the  positive  value  of  each  square  root,  the  above  is  not  a 
true  equation. 

Authorities  differ  as  to  whether  it  is  allowable  in  such  instance  to 
choose  the  negative  value  for  one  of  the  square  roots.  It  seems  more 
consistent  to  adhere  to  the  signs  expressed  in  the  given  equation.  If  this 
rule  is  followed,  the  above  equation  has  no  solution. 

EXERCISE   40 

Solve  the  following  equations  ;  verify  each  root : 
1.    Vx  +  5  +  2  =  5.  2.    Va  +  7  —  Vx=  1. 


3.    Vx2  +  ±x  _3-Va2  +  a  +  6:=0. 
4.    -^  +  11  +  4  =  7.  8.    V^-WS+8  =  -    12 


Vtf-f-8 


5.  Var*  — 11  -f  1  =  #.  /-  ,  1 1 

V  a?  4- 1 1  _  V  if  + 19 

6.  Vx  -  28  =  14  -  Vx.  VS-3        Vs%2 


/-  ,     /Tk 12  _     V4a;  +  5  +  Va;  +  3 

7.    Vif+VlO— if  =  —  .    10.    —      n       ' ! —  —  5. 

VlO  —  x  Vi  a  +  0  —  V»  +  3 


122  ALGEBRA 


ii     3y2  a?  +  4  =  3V2a?  +  2_        a      Va  +  #  +  Va  —  #_  1 
V2  a;  V2  #  -|- 1  Va  +  #  —  Va  —  x     * 


13-    Va?  +  m+  Va?  —  w  =  V2  m  —  ri  +  3  #. 


14.   Va  —  y-\-  V&  —  y=  Va~^ 


15.    V2s  +  3-V3s  +  3=-Vs-10. 
16.   A/r+Wf^"=  1  +  x.  19-    Va2-5a;-8  =  V.r-4. 


17.   x\U-l- V#  =  a.  20     Vb2 +  x  +  Vc2 +  x  =b^ 

■y/b2  +  x  —  Vc-  -\-x      c 
l8     V6-f  a?  +  V^  =  & 

V&  -\-x  —  Va 


VII.   IMAGINARY  NUMBERS 

196.  If  a  number  involves  an  indicated  even  root  of  a  nega- 
tive number,  it  is  called  imaginary.  Such  numbers  depend 
upon  a  new  unit,  V—  1  or  (—  1)*  ;  as  V—  2,  V  —  3. 

197.  An  imaginary  number  of  the  form  V—  a  is  called  a 
pure  imaginary  number,  and  the  sum  of  a  real  and  an  imagi- 
nary is  called  a  complex  number ;  as  a  +  b  V—  1. 

198.  Meaning  of  a  Pure  Imaginary  Number. 

If  Va  is  real,  we  define  Va  as  an  expression  such  that, 
when  raised  to  the  second  power,  the  result  is  a. 

To  find  what  meaning  to  attach  to  a  pure  imaginary  number, 
we  assume  the  above  principle  to  hold  when  Va  is  imaginary* 

Thus,  V— 2  means  an  expression  such  that,  when  raised  to 
the   second   power,   the   result   is  —  2 ;   that   is,  (^ 
(_2^)2=-2. 

In  like  manner,  ( V^l)2  =  (-  1*)2=  - 1 ;  etc. 


IMAGINARY   NUMBERS  123 

OPERATIONS   WITH   IMAGINARY   NUMBERS 

199.    By  §  198,  (V^)2  =  (-5*)2=--5.  (1) 

Also,  (V5V^l)2=(V5)2(V^l)2=5(-l)  =  -5,  (2) 

or  (V=5)2=(5*)s  •(-l^)2=5(-l)  =  -5. 

From  (1)  and  (2),  (V^5)2=  ( V5 V^)2. 

Whence,  V11^  =  -y/5V-l,  or  5*(- 1)*. 

Then,  every  imaginary  square  root  can  be  expressed   as  the 
product  of  a  real  number  by  V— 1.     It  is  advisable  to  reduce 
every  imaginary  to  this  form   before  performing   the  indicated 
operations. 
V—  1  is  called  the  imaginary  unit ;  it  is  often  represented  by  i. 

In  all  operations  with  imaginary  numbers,  it  is  advisable  to 
reduce  the  number  to  the  form  a  +  bi  where  a  and  b  are  real. 

Ex.    Add  V^4  and  V-36. 

V^4  =  2*\  V-  36  =  6  i. 

2t  +  6 i  =  8 *\  or  8v^l,  or  8  (-  l)i 

The  Powers  of  » :  V  —  1  =  i1 ; 

(V"=t:)3--v^i=^; 

Note  that  the  even  powers  of  i  are  real,  the  odd  powers 
imaginary,  the  fifth  power  like  the  first  power,  the  sixth 
power  like  the  second,  etc. 

Ex.  1.    Multiply  V-^2  by  V^S. 

vr^2  =  iv/2,    v^3  =  iV3. 

(iV2)  (iVS)  =  <*V5=-  \/S,  or  -(€)*. 


124  ALGEBRA 

Ex.  2.    Divide  (-40)2  by  (-5)'. 

(-  40)4  =  i  (40)2,  (-  §)♦  =  i  (5)* 

-^— %  =  (8)*  =  2  (2)*,  or  2 v2. 
EXERCISE  41 

Simplify  the  following : 

i.  V^l^+V^l. 

2.  2V:=^9  +  4V^25-3V^36. 

3.  2V::r3-3V^27  +  5V:=:12. 

4.  7  V^a2  -  3  V-  49  a2  -  2  V^4a2. 

5.  iV^8  +  iV^32-lV^162. 

6.  Add2  +  V:^to3-2vr=:27. 


7.  From  8-6  V-  121  subtract  5  +  2  V-169. 

8.  From  a  -  V26-62-l  take  6  -  V2  a  -  a2  - 1. 
Multiply  the  following : 

9.  V^byV^T.  11.    -V^Wby  V^~6. . 
10.  V^by  V-144.  12.    _V-432by-V:=~75. 

13.  V  —  a2,  V  —  52,  an d  —  V  —  c2. 

14.  2+V-3by  3-4V:r3. 

15.  5-2V-Tby  4-3i. 

16.  4t"VaJ  —  SiVy  by  9i'V#  +  V  —  2/. 
Expand  the  following : 

17.  (2-V^3)2.  18.    (3V^2  +  2V":^3)2. 

19.  (2V^4-3V^)(2V^-3V'^7). 

20.  (a  — V— 6)8. 


IMAGINARY   NUMBERS  125 

Divide  the  following: 

21.  V^^byV^.  23.    -Vl^by-V^ 

22.  V—  54  by  —  V—  3.  24.    —  V—  96a6  by  V—  Sab. 
25.   6^V6~VS84  by  -V^H 

GRAPHICAL  REPRESENTATION   OF   IMAGINARY   NUMBERS 

200.  Let  0  be  any  point  in  the  straight  line  XX1. 
We  may  suppose  any  positive  real 

number,   +  a,  to  be  represented    by  /^*\ 

the  distance  from  0  to  A,  a  units  to  2     A'   -&    0+^    A     x 

the  right  of  O  in  OX. 

Then  any  negative  real  number,  —  a,  may  be  represented 
by  the  distance  from  0  to  A',  a  units  to  the  left  of  0  in  OX'. 

201.  Since  —  a  is  the  same  as  (-f-a)x(- 1),  it  follows 
from  §  200  that  the  product  of  +  a  by  —  1  is  represented  by 
turning  the  line  OA,  which  represents  the  number  +  a, 
through  two  right  angles,  in  a  direction  opposite  to  the  motion 
of  the  hands  of  a  clock. 

Then,  in  the  product  of  any  real  number  by  —  1,  we  may 
regard  —  1  as  an  operator  which  turns  $he  line  which  repre- 
sents the  first  factor  through  two  right  angles,  in  a  direction 
opposite  to  the  motion  of  the  hands  of  a  clock. 

202.  Graphical  Representation  of  the  Imaginary  Unit  i  (§  196). 
By  the  definition  of  §  198,  —  1  =  1  x  ii 
Then,  since  one  multiplication  by  i,  fol- 
lowed by  another  multiplication  by  i,  turns 
the   line  which  represents  the   first  factor 

X- 1 — ^ 1 2   through   two   right    angles,  in    a   direction 

opposite  to  the  hands  of  a  clock,  we  may 
regard  multiplication  by  1  as  turning  the 
line  through  one  right  angle,  in  the  same 
direction. 
Thus,  let  XX  and  YY'  be  straight  lines  intersecting  at  right  angles 

at  O. 


Y 

•B 

c- 

+ai 

+i 

S  . 

nr 

0-f-a    A 

-1 

c'. 

-ai 
-B' 

f 

126 


ALGEBRA 


Then,  if  +  a  be  represented  by  the  line  OA,  where  A  is  a  units  to  the 
right  of  0  in  OX,  +  ai  may  be  represented  by  OB,  and  —  ai  by  OB', 
where  B  is  a  units  above,  and  B'  a  units  below,  O,  in  77'. 

Also,  +  i  may  be  represented  by  OC,  and  —  i  by  OC,  where  C  is  one 
unit  above,  and  C  one  unit  below,  0,  in  77'. 


203.    Graphical  Representation  of  Complex  Numbers. 

We  will  now  show  how  to  represent  the  complex  number 


a  +  bi. 
Y 

■  B 


rO 


,V 


*>A 


<v 


Let  XX  and  77'  be  straight  lines  intersecting 
at  right  angles  at  0. 

Let  a  be  represented  by  OA,  to  the  right  of  O, 
f    if  a  is  positive,  to  the  left  if  a  is  negative. 

Let  bi  be  represented  by  OB,  above  0  if  b  is 
positive,  below  if  b  is  negative. 
Draw  line  AC  equal  and  parallel  to  OB,  on  the  same  side  of  XX  as 
05,  and  line  OC. 

Then,  00  is  considered  as  representing  the  result  of  adding  bi  to  a  ; 
that  is,  OC  represents  the  complex  number  a  +  bi. 

The  figure  represents  the  case  in  which  both  a  and  b  are  positive. 


As  another  illustration,  we  will  show  how  to  represent  the 
complex  number  —  5  —  4  i. 

Lay  off  OA  5  units  to  the  left  of  O  in  OX, 
and  OB  4  units  below  0  in  YY'. 

Draw  line  ^40  below  XX,  equal  and  parallel 
to  OB,  and  line  OC. 

Then,  06y  represents  —  5  —  4  i. 

The  complex  number  a  +  bi,  if  a  is  positive  and 
b  negative,  will  be  represented  by  a  line  between 
OX  and  07' ;  and  if  a  is  negative  and  b  positive,  by  a  line  between  07 
and  OX. 


7 

x- 

,  A      -r> 

0 

I 

*> 

c^. 

/X 

C 

v>' 

B 

r' 

EXERCISE  42 

Represent  the  following  graphically  : 
i.   3  I  2.    —  61  3.   4  +  i. 

5.   2-5i.  6.    -5-:W. 


4.    _l  +  2  t\ 

7.    -7  +  4t. 


IMAGINARY   NUMBERS 


127 


204.  Graphical  Representation  of  Addition. 

We  will  now  show  how  to  represent  the  result  of  adding  b  to 
<i,  where  a  and  b  are  any  two  real,  pure  imaginary,  or  complex 
numbers. 

Let  the  line  a  be  represented  by  OA,  and  the  line 
h  by  OB. 

Draw  the  line  A  C  equal  and  parallel  to  OB,  on 
the  same  side  of  OA  as  OB,  and  the  line  OC. 

Then,  0(7  is  considered  as  representing  the  result 
of  adding  b  to  a  ;  that  is,  00  represents  a  +  b. 

The  method  of  §  20*3  is  a  special  case  of  the  above. 

If  a  and  b  are  both  real,  B  will  fall  in  OA,  or  in  AO  produced 
through  O. 

The  same  will  be  true  if  a  and  b  are  both  pure  imaginary. 

If  one  of  the  numbers,  a  and  b,  is  real,  and  the  other  pure  imaginary, 
the  lines  OA  and  OB  will  be  perpendicular. 

As  another  illustration,  we  will  show  how  to  represent 
graphically   the   sum   of    the   complex   numbers    2  —  5  i  and 

-4  +  3*. 

The  complex  number  2  —  5  i  is  represented  by  the 
line  OA,  between  OX  and  OF. 

The  complex  number  —  4  4-  3  i  is  represented  by 
the  line  OB,  between  0  Y  and  OX'. 

Draw  the  line  BG  equal  and  parallel  to  OA,  on 
the  same  side  of  OB  as  OA,  and  the  line  OC 

Then,  the  line  OC  represents  the  result  of  adding 
-  4  +  3  i  to  2  —  5  i. 

205.  Graphical  Representation  of  Subtraction. 

Let  a  and  b  be  any  two  real,  pure  imaginary,  or  complex 
numbers. 

Let  a  be  represented  by   OA,  and  b  by  OB  j  By 
and  complete  the  parallelogram  OB  AC. 

By  §  204,  0^4  represents  the  result  of  adding 
the  number  represented  by  OB  to  the  number 
represented  by  OC. 

That  is,  if  b  be  added  to  the  number  represented 
by  OC,  the  sum  is  equal  to  a ;  hence,  a  —  b  is 
represented  by  the  line  OC. 


128  ALGEBRA 

EXERCISE  43 

Represent  the  following  graphically  : 
i.    The  sum  of  4  i  and  3  —  5i. 

2.  The  sum  of  —  5  i  and  —  1  +  6  i. 

3.  The  sum  of  2  +  4  *  and  5  —  3  i. 

4.  The  sum  of  —  6  -f  2  1  and  —  4  —  7  i. 

5.  Represent  graphically  the  result  of  subtracting  the  second 
expression  from  the  first,  in  each  of  the  above  examples. 

VIII.    QUADRATIC   EQUATIONS 

206.  A  quadratic  equation  is  an  equation  in  which  the 
highest  power  of  the  unknown  number  is  the  second. 

207.  The  first  power  of  the  unknown  number  may  or  may 
not  appear.  If  the  equation  does,  not  contain  the  first  degree 
of  the  unknown,  the  roots  are  of  the  same  absolute  value  but  of 
different  sign.  E.g.  x2  =  a2 ;  then,  (x  +  a)(x  —  a)  =  0,  or  x  =  a, 
x  =  —  a. 

The  equation  may  also  be  solved  by  extracting  the  square 
root  of  each  member  of  the.  equation,  whence,  x  =  ±  a. 

208.  If  the  equation  contains  both  the  first  and  second 
powers  of  the  unknown,  the  first  member  must  be  reduced  to 
the  form  a2  +  2  ab  +  b2  before  extracting  the  square  root.  Such 
transformation  of  the  equation  is  called  completing  the  square. 

209.  A  quadratic  equation  containing  the  first  and  second 
powers  of  the  unknown  number  is  called  an  affected  quadratic. 
An  equation  containing  the  second  degree  only  of  the  unknown 
number  is  a  pure  quadratic. 

AFFECTED    QUADRATIC   EQUATIONS 

210.  First  Method  of  Completing  the  Square. 

By  transposing  the  terms  involving  x  to  the  first  member, 
and  all  other  terms  to  the  second,  and   then  dividing  both 


QUADRATIC    EQUATIONS  '  129 

members  by  the  coefficient  of  x2,  any  affected  quadratic  equa- 
tion can  be  reduced  to  the  form  x2  -f-  px  —  q. 

We  then  add  to  both  members  such  an  expression  as  will 
make  the  first  member  a  trinomial  perfect  square  (§  103,  II)  ; 
an  operation  which  is  termed  completing  the  square. 

Ex.    Solve  the  equation  x2  -f  3  x  =  4. 

A  trinomial  is  a  perfect  square  when  its  first  and  third  terms  are  per- 
fect squares  and  positive,  and  its  second  term  plus  or  minus  twice  the 
product  of  their  square  roots  (§  103,  II). 

Then,  the  square  root  of  the  third  term  is  equal  to  the  second  term 
divided  by  twice  the  square  root  of  the  first. 

Hence,  the  square  root  of  the  expression  which  must  be  added  to 
x2  +  3  x  to  make  it  a  perfect  square  is  3  x  -f-  2  jc,  or  |. 

Adding  to  both  members  the  square  of  |,  we  have 

X2  +  3  X  +  (|)2  =rr*  +  f  =a  \5-. 

Equating  the  square  root  of  the  first  member  to  the  ±  square  root  of 
the  second,  we  have 

x  +  |  =  ±  | . 

Transposing  f ,  x  =  —  f  +  §  or  —  §  —  §  =  1  or  — -  4. 

We  then  have  the  following  rule : 

Reduce  the  equation  to  the  form  x2  +  px  =  q. 

Complete  the  square,  by  adding  to  both  members  the 
square  of  one-half  the  coefficient  of  a?. 

Equate  the  square  root  of  the  first  member  to  the 
±  square  root  of  the  second,  and  solve  the  linear  equa- 
tions thus  formed. 

21 1.  The  objection  to  the  method  of  §  210  is  that  in  dividing 
by  the  coefficient  of  x2,  or  in  adding  the  square  of  one-half  the 
coefficient  of  x,  fractions  which  make  the  solution  cumbersome 
may  be  introduced. 

212.  If  the  coefficient  of  x2  is  a  perfect  square,  it  is  some- 
times convenient  to  complete  the  square  directly  by  the  prin- 
ciple stated  in  §  210;  that  is,  by  adding  to  both  members  the 
square  of  the  quotient  obtained  by  dividing  the  coefficient  of  x 
by  twice  the  square  root  of  the  coefficient  of  x2. 


130  ALGEBRA 

Ex.    Solve  the  equation  9  x2  —  5  x  =  4. 

5  5 

Adding  to  both  members  the  square  of ,  or  -, 

2x3.      6 

9*2-5*  +  (5)2  =  4  +JS  =  W- 

Extracting  square  roots,  3  a:  —  §  =  ±  JB3- 

Then,  3  se  —  |  ±  a^  =  3  or  f ,  and  x  =  1  or  —  j. 

213.    Second  Method  of  Completing  the  Square. 
Every  affected  quadratic  equation  can  be  reduced  to  the  form 
ax2  +  bx  -|-  c  =  0,  or  ax2  -\-bx  =  —  c. 

Multiplying  both  members  by  4  a,  we  have 

4  a2x2  +  4  a&ac  =  —  4  ac. 

We  complete  the  square  by  adding  to  both  members  the  square  of 

-t^-  (§  212),  or  b. 
2  x  2a 

Then,  4  a2£2  +  4  afoc  +  ft8  =  62  -  4  ac. 


Extracting  square  roots,  2  ax  +  6  =  ±  V6-  —  4  ac. 


Transposing,  2ax  =—  b  ±  Vb2  —  4  ac. 


Whence,  x  =  -  &  ±  V6*  -  4  ac.  (1) 

-  a 

We  then  have  the  following  rule : 

Reduce  the  equation  to  the  form  aac2  +  bx  =—  c. 

Multiply  both  members  by  four  times  the  coefficient 
of  as*,  and  add  to  each  the  square  of  the  coefficient  of 
x  in  the  given  equation. 

The  advantage  of  this  method  over  the  preceding  is  in 
avoiding  fractions  in  completing  the  square. 

This  method  is  sometimes  called  the  Hindoo  Method. 

The  result  of  the  solution  of  ax2  4-  bx  +  c  =  0  may  be  used  as 
a  formula  for  solving  any  quadratic  equation.  Before  apply- 
ing the  formula  the  equation  must  be  reduced  to  the  form 

ax2  +  bx  -f  c  =  0. 


QUADRATIC   EQUATIONS  131 

Ex.    Solve  2  x2  —  7  x  =  —  3. 

2  .x-2  -  7  x  +  8  =  0. 
Here  a  =  2,  £>=—  7,  c  =  3  ;  substituting  in  (1), 


•  =  3  or  -. 

22  4  2 

EXERCISE   44 

Solve  by  the  first  method :     (Verify  each  result.) 
i.    ar- 12^4- 32  =  0.  6.    *2  +  *_30  =  0. 


z2  +  7  z- 30  =  0.  7.    6  z2  4-4  =  -11  z. 

4^-Tf  — 8.  8    4*2-3*  =  7. 

16x2-8x-35=0. 

9.    ?_2_^_M  =  o. 
3  m2 -26  =  9  m2 -80.  3      2       6 

3a;2  2a?-6  =  1 2_ 


x2  —  7#4-6       x  —  6 


Solve  by  second  method  :     (Verify  each  result.) 
ii.    (3fc4-2)(2&H-3)  =  (fc-3)(2fc-4). 


12.    30-  3° 


07         iC-f  1 


13.  Vm  -f  2  4-  V3  m  +  4  =  8. 

14.  (y-3y-(y  +  2)*  =  -65. 

15.  V5  4-  x  4-  V5  —  x  =  - 


16. 

d-2 

iVote  i  :    In  solving  equations  involving  fractions  or  radicals  reject  any 
root  which  does  not  satisfy  the  given  equation. 


132  ALGEBRA 

Solve  by  means  of  the  formula  in  §213,  (1):  (Verify  each 
result.) 

17.  3x2-2x  =  A0.  5       13        1 

18.  9x2  +  18x=-8.  .    6*     9z"     1S 

1  1  a2- 17 

20. 


x  +  3     x-5     x2-2x-15 

21.  y~c     y  +  c^if-Bc2^ 
y  +  c     y  —  c     y2  —  <? 

1  Iz  15 

22. +  ■ 


-2      24(2  +  2)      z2-4 

1      ,1  ,      1      ,  1      n 

23.    -H 1 |--  =  0o 

#  +  <z      a     #-+-  6      6 

24.  #=F*  +  £gtf.2     Solve  fori. 

a;  —  2      #4-2     x  —  2  ^      /a      XT  .    0  v 

2«. ?__ =— 1.    (See  Note  2.) 

26.    ^  +  1  .i_fl  +  2  .a?  +  3=_3 

#  — 1      #  —  2      x  —  3 

Note  2 :  In  solving  fractional  equations  containing  improper  fractions 
the  operations  are  greatly  simplified  by  reducing  the  fractions  to  mixed 
numbers  and  then  combining  the  integers  thus  obtained. 

Note  3 :  An  equation  is  said  to  be  in  the  quadratic  form  when  it  is 
expressed  in  three  terms,  two-  of  which  contain  the  unknown  number, 
and  the  exponent  of  the  unknown  number  in  one  of  these  terms  is  ticicc  its 
exponent  in  the  other  ;  as, 

xe  _  0  xz  _  iq  .  3*  +  xi  _  72  =  0  ;  etc. 

Equations  in  the  quadratic  form  may  be  readily  solved  by 
the  rules  for  quadratics. 

1.    Solve  the  equation  x6  —  6  Xs  =  16. 
Completing  the  square  by  the  rule  of  §  210, 

x6  -  0  x3  +  9  =  10  +  (.)  =  25. 


QUADRATIC    EQUATIONS  133 

Extracting  square  roots,      x8  —  3  =  ±  5. 

Then,  x3  =  3  ±  5  =  8  or  -  2. 

Extracting  cube  roots,  x  —  2  or  —  v2. 

There  are  also  four  imaginary  roots,  which  may  be  found  by  the 
method  of  §§  110;  213,  (1). 

Solve  the  equation  2  x  -f  3  V#  =  27. 

_  i 

Since  Vx  is  the  same  asx^,  this  is  in  the  quadratic  form. 

Multiplying  by  8,  and  adding  32  to  both  members  §  (213), 

16  x  +  24  y/x  +  9  =  216  +  9  =  225. 
Extracting  square  roots,  4  Vx  +'8=  ±  15. 
Then,  4Vx  =  -  3  ±  15  =  12  or  -  18. 

Whence,  Vx  =  3  or  — f,  and  sc  =  9  or  *£. 

EXERCISE   45 

Solve  the  following  equations  and  verify  each  root: 
i.   3a2-4a  =  4.  (      4.   2*  =  10-£2+5*. 

„    n   2     4*        o    2  ,  oo  5-   6v2— 14^  =  9  v  —  22. 

2.  7  ar  —  17  »  =  2  x2  +  22. 

6     ^~  3  .  x  +  5    __      8 

3.  42/2  +  92/-13  =  0.  '    a;  +  3      »-ll~       7* 

3  ra  -  7  1  2  -  ra 


7- 


m(m-f-2)      3(m-2)      ra2-4" 


M_  =  4.  9.    thJ+EH^^BL 


, .  21 

io.    l  +  2V3o;  +  2=    /o — 7-r,' 
V3  #-f2 

ii.    VaT^  +  2V2~&  = 


12. 


11  20  3  a 


3v-4a     2v-\-a     6v2—  5  av  —  4  a2      5  a' 


134  ALGEBRA 

a2-b* 


13-    -=-  + 


b  —  z         ab 


2  x 

15.  x*-2  ax  -f  6  b2  =  3  a2  +  7  a&  -  5  bx. 

16.  (2  a  -  b  +  5  c)  a2  +  (b  -  a  +  4  p)a  +  2  &  -  3  a-  c  =  0. 

17.  — 1 |--H —  +-  =  0.     Solve  ford. 

d  +  a      a     d+b      b 

Q     Vm2  -f  x2  +  Vm2  —  x2     m 

Io.      -  = . 

Vm2  -f  x2  —  -y/m2  —  x2  x 

19.  x4-7a2  +  12:=0.  21.   6  a-2 -11  a-1  =  35. 

20.  a6  — 7#3=8.  22.    x?  —  x*  —  6  =  0. 
23.    ar3  -  35  x§  =  -  216. 


24.  x2  +  2  a  +  10  +  VV  +  2  x  +  10  =  30. 

25.  #24-3a#  —  53  a2  =  2  a# -f- 3  a. 

26.  a?-* -29  a;-*  =-100. 


27.  .T2  +  14Var>+7a?-26  =  58-7a. 

28.  5  (a +  2)*  + (a? +  2)  =  36. 

29.  (a2  +  4a4-2)2  =  31+2(a2  +  4a  +  2). 

30.  a4-8arJ  +  10#2  +  24#-315  =  0. 

31.  What  number  is  that  to  which  if  you  add  its  square  the 
sum  will  be  42  ? 

32.  A  rectangular  field  is  40  rods  longer  than  it  is  wide.  By 
doubling  its  length  and  decreasing  its  width  by  15  rods,  the  area 
is  unchanged.     Find  dimensions  of  the  field. 

32-  The  difference  between  two  numbers  is  7,  and  the  dif- 
ference between  their  cubes  is  1267.     Find  the  numbers. 

34.  The  denominator  of  a  fraction  is  3  more  than  its  numera- 
tor and  by  adding  the  fraction  to  its  reciprocal  the  sum  is 
2^.     What  is  the  fraction  ? 


QUADRATIC   EQUATIONS  135 

35.  There  is  a  number  consisting  of  two  digits  whose  sum 
is  11.  If  from  the  number  3  times  the  product  of  the  digits 
is  subtracted,  the  remainder  will  equal  the  sum  of  the  digits. 
Find  the  number. 

36.  A  man  sells  goods  for  $120,  gaining  a  per  cent  equal 
to  I  the  cost  of  the  gooo^s.     What  was  the  cost  of  the  goods  ? 

37.  A  picture  13  inches  by  8  inches  is  surrounded  by  a 
frame  of  uniform  width  whose  area  is  162  square  inches. 
Find  the  width  of  the  frame. 

38.  A  man  put  $  2400  in  a  savings  bank  which  paid  inter- 
est semiannually.  At  the  end  of  a  year  he  found  that  he  had 
to  his  credit  $  2496.96.     What  interest  did  the  bank  pay  ? 

39.  A  number  of  people  plan  an  excursion  which  is  to  cost 
them  $  30.  It  is  found  later  that  3  of  the  party  cannot  go, 
which  increases  the  cost  50  cents  to  each  member.  How  many 
are  there  in  the  party  and  what  did  each  one  pay  ? 

40.  A  and  B  start  together  for  a  6-mile  walk.  A's  rate  per  hour 
is  \  mile  more  than  B's,  and  he  finds  he  can  reach  his  destina- 
tion in  24  minutes  less  time  than  B.    What  is  the  rate  of  each  ? 

41.  An  open  rectangular  box  is  8  inches  high.  Its  length 
is  4  more  than  its  width.  Its  volume  is  768  cubic  inches. 
Find  its  inside  dimensions. 

42.  In  a  given  circle  APB,  a  perpen- 
dicular DP,  dropped  from  a  point  P  in 
the  circumference  to  the  diameter  AB, 
is  a  mean  proportional  between  the  seg- 
ments, AD  and  DB,  of  the  diameter.  If 
the  radius  of  the  circle  is  12  and  DP  is 
2V5,  how  far  is  D  from  B? 

43.  An  open  rectangular  box  5  inches  deep  (inside  measure) 
is  made  of  1-inch  lumber.  Its  length  is  1  inch  less  than  twice 
its  width.  The  difference  between  the  volumes  when  inside  and 
outside  measurements  are  taken  is  271  cubic  inches.  How  much 
sheet  metal  will  be  needed  for  lining  the  sides  and  bottom  of 
the  box  ? 


136  ALGEBRA 

44.  Two  lines  AB  and  CD  intersect  at  0  in  such  a  manner 
that  AO-OB=CO-  OD.  If  CD  =  U,AO  =  15,  and  AB  =  18, 
find  CO. 

45.  A  has  a  lease  on  a  square  room.  He  sublet  to  B  a  part 
10  feet  wide  along  one  entire  side  of  the  room,  at  a  rental  of 
$  160  per  month.    The  part  of  the  room  retained  by  A  contained 

704  square  feet.  How  much  rental  per  square 
foot  did  B  pay  ?  Explain  your  negative  roots. 
46.  A  tangent,  PT,  to  a  circle  is  a  mean  pro- 
portional between  the  whole  secant  PD  and  the 
external  segment  PE.  If  PT  is  12,  the  radius 
5,  and  PD  passes  through  the  center,  find  PE. 

47.  The  upper  base  and  the  altitude  of  a  trapezoid  are  equal, 
the  lower  base  is  20  and  the  area  is  112.    Find  the  upper  base. 

48.  The  length  of  a  rectangle  is  V2  more  than  the  side  of  a 
given  square,  and  its  breadth  is  V2  less  than  a  side  of  the 
same  square.  The  area  of  the  rectangle  is  1.  Find  the  di- 
mensions of  the  rectangle  correct  to  three  decimal  places. 

THEORY   OF   QUADRATIC   EQUATIONS 

214.    Number  of  Roots. 

A  quadratic  equation  cannot  have  more  than  two  dif- 
ferent roots. 

Every  quadratic  equation  can  be  reduced  to  the  form 
ax2  +  bx  +  c  =  0.  * 

If  possible,  let  this  have  three  different  roots,  n,  ft,  and  r$. 

Then,  an2  +  brx  +  c  =  0,  (1) 

ar22  +  br2  +  c  =  0,  (2) 

and  ars2  +  br3  +  c  =  0.  (3) 

Subtracting  (2)  from  (1),        a{rx2  -  r22)  +  &(r*  -  r2)  =  0. 

Then,  a(n  +  r%)  (n.  —  r2)  4-  &Oi  —  r2)  =  0, 

or,  (n  —  r2)  (ari  +  ar2  +  6)  —  0. 

Then,  by  §  110,  either        ri  —  r2  =*  0,  or  ar\  -f  oft  +  ?>  =  0. 

But  ri  —  r2  cannot  equal  0,  for,  by  hypothesis,  r*i  and  r2  are  different. 


QUADRATIC   EQUATIONS 


137 


Whence, 


ari  +  ar-2,  +  6  =  0. 


(4) 


Iii  like  manner,  by  subtracting  (3)  from  (1),  we  have 

ari  +  an  +  6  =  0.  (5) 

Subtracting  (5)  from  (4),        ar2  —  an  =  0,  or  r2  —  n  =  0. 

But  this  is   impossible,  for,  by  hypothesis,  r2   and  n   are  different ; 
hence,  a  quadratic  equation  cannot  have  more  than  two  different  roots. 

215.    The  graphs  of  quadratic  equations  can  be  readily  con- 
structed by  the  method  used  in  §§  44-48. 


Construct  the  graph  of 


x- 


■6  =  0. 


Placing  the  first  member  of  the  equation  equal  to  y,  we  have 
x2  —  x  —  6  =  y. 


a) 


(2) 


Assigning  values  to  x,  we  obtain   corresponding  values  of  y.     For 
example,  Substituting  x  =  0  in  (2),  we  have  y  =  -  6, 

Substituting  x  =  2  in  (2),  we  have  y  =  —  4,  etc. 

x2  —  x  —  6  =  ?y 


2/ 

-6 

-6i 

(-1) 

-6 

-  51  (-B) 

-4 

(O) 

-  2J  (D) 

0 

(E) 

6 

-51 

(G) 

-4 

(H) 

0 

W 

6 

Y 

\ 

\ 

\ 

\ 

X' 

K) 

O 

/E 

>■ 

Yc 

H 

c 

/P 

A 

Y' 

X2  _  x  _  e  s-  0, 
(£-3)(x  +  2)  =0, 

x  =  3  or 


Solving 
or 
we  have,  x  —  3  or  —  2. 

These  values,  x  =  3,  x  =  —  2,  are  the  abscissas  of  the  points  where  the 
curve  crosses  the  x-axis,  the  curve  showing  in  a  graphical  way  why  a 
quadratic  equation  has  two  roots. 


138  ALGEBRA 

The  graph  of  every  equation  of  the  form  x2  +  px  —  q  =  0  or 
ax2  -f  bx  -f-  c  =  0  is  a  curve  of  the  above  form  and  is  called  a 
parabola. 

216.    Sum  of  Roots  and  Product  of  Roots. 

Let  rx  and  r2  denote  the  roots  of  ax2  -f-  bx  +  c  =  0. 


By  §  213,  (1),  n  =  -ft  +  V5'-4qc   and  ^  =  -  b  -  Vb2  -  4  flC. 

2  a  2  a 

Adding  these  values,  n  -f  ra  ==  — — -  =  —  -• 

2  a  a 

Multiplying  them  together, 

nn  =  ^-{b^-iac)  (    103   j   =  4_ae  =  c . 
4  a2  4  a2      a 

Hence,  if  a  quadratic  equation  is  in  the  form  ax2  -f  bx 
+  c  =  0,  the  sum  of  the  roots  equals  minus  the  coefficient 
of  x  divided  by  the  coefficient  of  a?2,  and  the  product  of 
the  roots  equals  the  independent  term  divided  by  the 
coefficient  of  a?2. 

217.   Formation  of  Quadratic  Equations. 

By  aid  of  the  principles  of  §  216,  a  quadratic  equation  may 
be  formed  which  shall  have  any  required  roots. 

For,  let  ri  and  r2  denote  the  roots  of  the  equation 

ax2  +  bx  +  c  =  0,  or  a2  +  —  +  -  =  0.  (1) 

a      a 

Then  by  §  216,  5  =  _  n  -  r2,  and  G-  =  nr2. 

a  a 

Substituting  these  values  in  (1),  we  have 

x2  —  r\X  —  r2x  +  7"ir2  =  0. 
Or,  (x  —  rx)  (x  —  r2)  =  0. 

Therefore,  to  form  a  quadratic  equation  which  shall  have 
any  required  roots, 

Subtract  each  of  the  roots  from  05,  and  place  the  prod- 
uct of  the  resulting  expressions  equal  to  zero. 

Ex.    Form  the  quadratic  whose  roots  shall  be  4  and  —  -J. 

By  the  rule,  («  -  4)  (x  +  J)  =  0. 

Multiplying  by  4,  {x  -  4)  (4  x  +  7)  =  0  ;  or,  4  x2  -  9  x  -  28  =  0. 


QUADRATIC    EQUATIONS  139 

DISCUSSION    OF    GENERAL   EQUATION 

218.   The  roots  of  a  quadratic  equation   may  take   several 
forms  : 

1.  The  roots  may  be  rational,  unequal,  of  the  same  sign. 

2.  The  roots  may  be  rational,  unequal,  of  opposite  sign. 

3.  The  roots  may  be  rational,  equal. 

4.  The  roots  may  be  irrational,  unequal. 

5.  The  roots  may  be  irrational,  equal. 

6.  The  roots  may  be  irrational  and  the  number  under  the 
radical  sign  negative. 

These  forms  and  the  causes  for  their  existence  are  at  once 
seen  when  one  considers  the  formula  in  §  213. 
By  §  213,  the  roots  of  ax2  +  bx  -f-  c  ==  0  are 

—  b  4-  V&2  —  1  ac       -,  —  b  —  -\/b2  —  4  ac 

r,  = :£— ■  and  r2  = 

1  2a  2a 

We  will  now  discuss  these  results  for  all  possible  real  values 

of  a,  b,  and  c. 

I.  b2  —  4  ac  positive. 

In  this  case,  i\  and  r2  are  real  and  unequal. 

II.  &*-4ac  =  0. 

In  this  case,  rx  and  r2  are  real  and  equal. 

III.  c  =  0. 

In  this  case,  the  equation  takes  the  form 

ax2  +  bx  =  0 :  whence  x  =  0  or 

a 

Hence,  the  roots  are  both  real,  one  being  zero. 

IV.  b  =  0,  and  c  =  0. 

In  this  case,  the  equation  takes  the  form  ax2  =  0. 
Hence,  both  roots  equal  zero. 

V.  b2  —  4  ac  negative. 

•    In  this  case,  rx  and  r2  are  imaginary  (§  196). 

VI.  6  =  0. 

In  this  case,  the  equation  takes  the  form         

ax2  -f  c  =  0 ;  whence,  a?  =  ±  \/ 


140  ALGEBRA 

If  a  and  c  are  of  unlike  sign,  the  roots  are  real,  equal  in 
absolute  value,  and  unlike  in  sign. 

If  a  and  c  are  of  like  sign,  both  roots  are  imaginary. 

The  roots  are  both  rational,  or  both  irrational,  according  as 
b2  —  4  ac  is,  or  is  not,  a  perfect  square. 

219.  It  is  evident  that  irrational  roots,  whether  real  or 
imaginary,  must  occur  in  conjugate  pairs. 

That  is,  in  an  equation  of  the  form  of  ax*  -f  bx  +  c  =  0,  where 
a,  b,  c  are  real,  if  one  root  is  of  the  form  k  -f  y7t  the  other 
must  be  k  —  ^/h  where  k  and  h  are  real. 

EXERCISE   46 

Find  by  inspection  the  sum  and  product  of  the  roots  of  the 
following : 

i.   x2  —  2  x  —  35  =  0.  5-   x2  -f  ax  —  bx  =  ab. 

2.  x2  +  15^  +  36  =  0.  6.   cdx2  +  d2x  =  c2x  +  cd. 

3.  2rJ  +  7a;-4  =  0.  7-   #2-2V2a-2  =  0. 
4-   5a2-13a  =  -6. 

8.  One  root  of  8  x2  —  2  a?  —  15  =  0  is  —  1^;  find  the  other. 

9.  One  root  of  6X2  +  11  x—  2  =  0  is  ^;  find  the  other. 

10.  One  root  of  2ar5-8a;2+2a4-12=0  is2;  find  the  others. 

11.  One  root  of  m3  —  7  m  +  6  =  0  is  —  3  ;  find  the  others. 

12.  If  rY  and  r2  are  the  roots  of  x2  -f  x  -f- 1  =  0,  what  does 
r2  -+-  r22  equal  ?     r?  -f-  ^3  ? 

Form  the  equations  whose  roots  are : 

13-   2,  3.  18.   a,  6  a. 

14.  —  1,  4.  19.   a  4-  V&,  a  —  V&. 

15.  i-3.  20.   2+V^3,  2-V^3. 

16.  -|,  —  £  21.   3c-d,  —  2c+5f7. 

17.  0,  -£. 


QUADRATIC   EQUATIONS 


141 


22. 


23- 


V2k-5Vg     y^2k±jyVgt 

2  '  2 

6,-1,0. 

Determine  by   inspection   the  nature   of   the   roots  of  the 
following : 

24.  x2  +  7  x  +  12  =  0.  30. 

25.  x2  +  8x  =  -16.  31. 

26.  £2  +  2a-l  =  0.  32. 

27.  ^  +  2a  +  3  =  0.  33- 

28.  2  a;2 +  7  a=3.  34- 

29.  4  a2  -16  =  0.  35- 


2a2  =  15z  +  18. 
x2  -  x  =  12. 
10a2-z  =  2. 
23  a?  -  6  =  7  a2. 
16  x2  +  24  x  +  9  =  0. 
5x2  +  3x  =  -2. 


GRAPHS 

220.    The  nature  of  the  roots  discussed  in  §  218  is  illustrated 
by  the  use  of  graphs : 


Y 

Fig   9 

Y 

1 

I 

Fig    "■ 

1 

y 

-8 

-9 

-8 

-5 

0 

7 

-5 

0 

7 

1 

X 

0 
1 
2 
8 
4 
5 
-2 
-3 

y 

1 
0 
1 
4 
9 

16 

9 

16 

\ 

X 

\ 

\ 

/ 

0 

\ 

/ 

\ 

/ 

\ 

/ 

i 

1 

x' 

O 

X 

2 

/ 

I 

3 

\ 

/ 

\ 

/ 

4 

\ 

/ 

\ 

/ 

5 

\ 

/ 

\ 

/ 

-1 

\ 

I 

\ 

1 

-? 

\ 

1 

\ 

3 

\ 

\ 

/ 

\ 

1 

X' 

\ 

/ 

f 

X 

V 

/ 

0 

y' 

xl 

Fig.  1.     #2_  2^-8  =  0 

62-4ac>0 


Fig.  2.     #2-2a?  +  l=0 
ft2  -  4  ac  =  O 


142 


ALGEBRA 


Fig 

3 

X 

y 

0 

3 

1 

2 

2 

3 

3 

6 

4 

11 

-  1 

6 

-2 

11 

Y 

Snip 
1111 

xj o x 

y' 


-3|  18 


3.     x*-23C  +  3  =  Q 

b2  -  4  ac  <  O 

In  Fig.  1,  the  curve  crosses  the  a>axis  at  points  whose  abscis- 
sas are  4,  —  2 ;  the  abscissas  of  these  points  being  the  values  of 
x  found  in  solving  the  equation.  In  Fig.  2,  the  intersection 
points  coincide  and  we  have  two  values  of  x  each  equal  to  1. 
In  Fig.  3,  the  curve  and  the  a>axis  do  not  coincide. 


EXERCISE    47 

Plot  the  curves : 

i.   f(x)  =  x2  +  6 x  +  8.  (§§  47,  220.) 

2.  f(x)  =  x2-6x  +  8.  4.  f(x)  =  x2-6x  +  9. 

3.  f(x)  =x2-9.  5.  f(x)  =  x2  +  2  x  +  4. 

221.  Many  problems  in  Physics  are  dependent  on  the  laws 
of  proportion  and  variation.  The  solution  of  such  problems 
is  often  obtained  more  readily  by  graphical  means  than  by 
algebraic  solution. 


QUADRATIC    EQUATIONS 


143 


Ex.  1.     Graphical  representation  of  a  direct  proportion. 
When  a  man  is  running  at  a  constant  speed,  the  distance 
which  he  travels  in  a  given  time  is  directly  proportional  to 

d      s 
his  speed.     The  algebraic  expression  of  this  relation  is  —  =  — , 

ord  =  ms.     (See  §161.)  d'2     * 

Now,  if  we  plot  successive  values  of  the  distance,  d,  which  correspond 
to  various  speeds,  a,  in  precisely  the  same 
manner  in  which  we  plotted  successive       ,  * 

values  of  x  and  y  in  §§  44-48,  we  obtain 
as  the  graphical  picture  of  the  relation 
between  s  and  d  a  straight  line  passing 
through  the  origin.     (See  Fig.  1.) 


This  is  the  graph  of  any  direct 
proportion. 


Fig.  l. 


Ex, 
P 


.2. 


r  \ 

•    i  x. 

j J i L— L-Ulpa 


Graphical  representation  of  an  inverse  proportion. 

The  volume  which  a  given  body 
of  gas  occupies  when  the  pressure 
to  which  it  is  subjected  varies  has 
been  found  to  be  inversely  propor- 
tional to  the  pressure  under  which 
the  gas  stands ;  we  have  seen  that 
the  algebraic  statement  of  this  re- 

V      P> 

lation  is  — -  =  tt* 

If  we  plot  successive  values  of  V  and 
P  in  the  manner  indicated  in  §§  44-48,  we 
obtain  a  graph  of  the  form  shown  in 
Fig.  2. 

This  is  the  graphical  representa- 
'  tion  of  any  inverse  proportion  ;  the 
curve  is  called  an  equilateral  hyper- 
•  bola. 

Ex.  3.  The  path  traversed  by  a 
'   falling  body  projected  horizontally. 


i    i  »    4 

5      6 

7      8    r 

Fig. 

2. 

V2      Px' 

or  V  = 

m 
P' 

V=l 

P  =  m. 

V=5, 

p_m 
5 

F=2, 

P  —  Wl 
2  ' 

r=6, 

P  -VI. 
(3 

V=3 

3 

V-=l, 

p_m 

7 

F=4 

p_m 
4  ' 

F=8, 

8 

144 


ALGEBRA 


When  a  body  is  thrown  horizontally  from  the  top  of  a  tower, 
if  it  were  not  for  gravity,  it  would  move  on  in  a  horizontal 
direction  indefinitely,  traversing  exactly  the  same  distance  in 
each  succeeding  second. 

Hence,  if  V  represents  the  velocity  of  projection,  the  horizontal  dis- 
tance, If,  which  it  would  traverse  in  any  number  of  seconds,  t,  would 
be  given  by  the  equation  H—  Vt. 

On  account  of  gravity,  however,  the  body  is  pulled  downward,  and 
traverses  in  this  direction  in  any  number  of  seconds  a  distance  which  is 
given  by  the  equation  8  =  \  gt2. 

To  find  the  actual  path  taken  by  the  body,  we  have  only  to  plot 
successive  values  of  II  and  S,  in  the  manner  in  which  we  plotted  the 
successive  values  of  x  and  ?/,  in  §§  44-48. 

Thus,  at  the  end  of  1  second  the  vertical  distance  Si  is  given  by 
#i  =  i  9  x  l2  =  \  g ;  at  the  end  of  2  seconds  we  have,  #2  =  J  g  x  22  =  \g  ; 
at  the  end  of  3  seconds,  #3  =  \  g  x  3'2  =  §  g ;   at  the  end  of  4  seconds, 

ft  =  . 

On  the  other  hand,  at  the  end  of  1  second 

we  have  H\  =  V  \  at  the  end  of  2  seconds, 
H2  —  2  V ;  at  the  end  of  3  seconds, 
H3=S  V;  at  the  end  of  4  seconds,  H4=4  V. 

If,  now,  we  plot  these  successive  values 
of  H  and  S,  we  obtain  the  graph  shown  in 
Fig.  3. 

This  is  the  path  of  the  body ;  it  is  a 
parabola.     (§  226,  Ex.  2.) 

Ex.  4.    GrapJi  of  relation  between 

the  temperature  and  pressure  existing 

within  an  air-tight  boiler  containing 

only  water  and  water  vapor. 

One  use  of  graphs  in  physics  is  to  express  a  relation  which 

is  found  by  experiment  to  exist  between  two  quantities,  which 

cannot  be  represented  by  any  simple  algebraic  equation. 

For  example,  when  the  temperature  of  an  air-tight  boiler 
which  contains  only  water  and  water  vapor  is  raised,  the  pres- 
sure within  the  boiler  increases  also;  thus  we  find  by  direct 
experiment  that  when  the  temperature  of  the  boiler. is  0°  centi- 
grade, the  pressure  which  the  vapor  exerts  will  support  a 
column  of  mercury  4.6  millimeters  high. 


Fig.  3, 


QUADRATIC    EQUATIONS 


145 


700 
000 


When  the  temperature  is  raised  to  10°,  the  mercury  column  rises  to 
9.1  millimeters  ;  at  30°  the  column  is  31.5  millimeters  long,  etc. 

To  obtain  a  simple  and  compact  picture  of  the  relation  between  tem- 
perature and  pressure,  we  plot  a  succession  of  temperatures,  e.g.  0°,  10°, 
20°,  80°,  40°,  50°,  60°,  70°,  80°,  90°, 
100°,  in  the  manner  in  which  we 
plotted  successive  values  of  x  in  §§  44- 
48,  and  then  plot  the  corresponding 
values  of  pressure  obtained  by  experi- 
ment in  the  manner  in  which  we 
plotted  the  ?/'s  in  §§  44-48  j  we  obtain 
the  graph  shown  in  Fig.  4. 

From  this  graph   we  can   find  at   200" 
once  the    pressure   which   will    exist 
within  the  boiler  at  any  temperature. 

For  example,  if  we  wish  to  know 
the    pressure  at    75°  centigrade,   we 
observe  where  the  vertical  line  which 
passes  through  75°  cuts  the  curve  and  then  run  a  horizontal  line  from  this 
point  to  the  point  of  intersection  with  the  line  OP. 

This  point  is  found  to  be  at  288  ;  hence  the  pressure  within  the  boiler 
at  75°  centigrade  is  288  millimeters. 


O    10     20    SO     40     50     00    TU 

Fig.  4. 


EXERCISE  48 
PROBLEMS   IN   PHYSICS 

i.  When  the  force  which  stretches  a  spring,  a  straight  wire, 
or  any  elastic  body  is  varied,  it  is  found  that  the  displacement 
produced  in  the  body  is  always  directly  proportional  to  the 
force  which  acts  upon  it ;  i.e.  if  e^  and  d2  represent  any  two 
displacements,  and  fx  and  f2  respectively  the  forces  which  pro- 
duce them,  then  the  algebraic  statement  of  the  above  law  is 

If  a  force  of  2  pounds  stretches  a  given  wire  .01  inch,  how 
much  will  a  force  of  20  pounds  stretch  the  same  wire  ? 

2.  If  the  same  force  is  applied  to  two  wires  of  the  same 
length  and  material,  but  of  different  diameters,  DY  and  D2,  then 


(i) 


146  ALGEBRA 

the  displacements  <lY  and  d2  are  found  to  be  inversely  propor- 
tional to  the  squares  of  the  diameters,  i.e. 

drw  () 

If  a  weight  of  100  kilograms  stretches  a  wire  .5  millimeter 
in  diameter  through  1  millimeter,  how  much  elongation  will 
the  same  weight  produce  in  a  wire  1.5  millimeters  in  diameter? 

3.  If  the  same  force  is  applied  to  two  wires  of  the  same 
diameter  and  material,  but  of  different  lengths,  lx  and  Z2,  then 
it  is  found  that  &J.& 

d2~k  ' 

Prom  (1),  (2),  and  (3)  and  §  164,  it  follows  that  when  lengths, 
diameters,  and  forces  are  all  different, 

d,    /■     h    A2  () 

If  a  force  of  1  ponnd  will  stretch  an  iron  wire  which  is 
200  centimeters  long  and  .5  millimeter  in  diameter  through 
1  millimeter,  what  force  is  required  to  stretch  an  iron  wire 
150  centimeters  long  and  1.25  millimeters  in  diameter  through 
.5  millimeter  ? 

4.  When  the  temperature  of  a  gas  is  constant,  its  volume 
is  found  to  be  inversely  proportional  to  the  pressure  to  which 
the  gas  is  subjected,  i.e.,  algebraically  stated, 

ZU5.  (5) 

V,    P,  w 

At  the  bottom  of  a  lake  30  meters  deep,  where  the  pressure 
is  4000  grams  per  square  centimeter,  a  bubble  of  air  has  a  vol- 
ume of  1  cubic  centimeter  as  it  escapes  from  a  diver's  suit.  To 
what  volume  will  it  have  expanded  when  it  reaches  the  surface 
where  the  atmospheric  pressure  is  about  1000  grams  per  square 
centimeter  ? 

5.  The  electrical  resistance  of  a  wire  varies  directly  as  its 
length  and  inversely  as  its  area.     If  a  copper  wire  1  centimeter 


QUADRATIC   EQUATIONS  147 

in  diameter  has  a  resistance  of  1  unit  per  mile,  how  many  units 
of  resistance  will  a  copper  wire  have  which  is  500  feet  long 
and  3  millimeters  in  diameter? 

6.  The  illumination  from  a  source  of  light  varies  inversely 
as  the  square  of  the  distance  from  the  source.  A  book  which 
is  now  10  inches  from  the  source  is  moved  15  inches  farther 
away.     How  much  will  the  light  received  be  reduced  ? 

7.  The  period  of  vibration  of  a  pendulum  is  found  to  vary 
directly  as  the  square  root  of  its  length.  If  a  pendulum  1  meter 
long  ticks  seconds,  what  will  be  the  period  of  vibration  of  a 
pendulum  30  centimeters  long  ? 

8.  The  force  with  which  the  earth  pulls  on  any  body  out- 
side of  its  surface  is  found  to  vary  inversely  as  the  square  of 
the  distance  from  its  center.  If  the  surface  of  the  earth  is 
4000  miles  from  the  center,  what  would  a  pound  weight  weigh 
15000  miles  from  the  earth  ? 

9.  The  number  of  vibrations  made  per  second  by  a  guitar 
string  of  given  diameter  and  material  is  inversely  proportional 
to  its  length  and  directly  proportional  to  the  square  root  of 
the  force  with  which  it  is  stretched.  If  a  string  3  feet  long, 
stretched  with  a  force  of  20  pounds,  vibrates  400  times  per 
second,  find  the  number  of  vibrations  made  by  a  string  1  foot 
long,  stretched  by  a  force  of  40  pounds. 

FACTORING 

In  Type  V,  §  103,  we  learned  to  transform  certain  trinomials 
into  Type  I,  §  103.  By  means  of  the  results  of  §  213,  we  are 
now  able  to  extend  this  method  to  expressions  not  readily  fac- 
tored by  the  simpler  processes. 

222.    Factoring  of  Quadratic  Expressions. 
A  quadratic  expression  is  an  expression  of  the  form 
ax2  -f  bx  -f-  c. 


148  ALGEBRA 

We  have, 

ax2  +  bx  +  c  =  a(x2  +  h~  +  -^ 
V  a      a) 

=«[-+i?+(fj-^i-] 


V        2a  2a        A        2a  2a        /' 

by  §  103,  I. 
But  by  §  213,  the  roots  of  ax2  +  bx  +  c  =  0  are 

__6       Vfr2-4qc  and  __b_       \/b2  -  4  ac 
2  a  2a  2  a  2a 

Hence,  to  factor  a  quadratic  expression  place  it  equal 
to  zero,  and  solve  the  equation  thus  formed. 

Then  the  required  factors  are  the  coefficient  of  a?2  in 
the  given  expression,  oc  minus  the  first  root,  and  x  minus 
the  second. 

Sometimes  the  expression  may  be  written  as  the  difference 
of  two  squares  and  the  method  of  §  103,  V,  used. 

Ex.  Factor  a4  + 1. 

x*  +  1  =  (x*  +  2  x2  +  1)  -  2  x2 
=  (x2  +  l)2-  <W2)2 
=  (x2  +  xV2  +  l)(x2  -  xV2  +  1). 

EXERCISE  49 

Factor  the  following : 

i.  a2 +  11  x  +  24.  8.  16  +  18  b-9  b\ 

2.  m2-m- 210.  9-  2  +  5p-25p2. 

3.  3a2-10a-8.  10.  49a2  +  28a-ll. 

4.  2  a2 -11  a +  15.  11.  a4 +  4. 

5.  28 -13  a -6  or*.    .  12.  a4+y. 

6.  8c2  +  8c-6.  13.  9tf4  +  22d*  +  25. 

7.  9a2-6a-8.  14.  16  a4  -  78  a262  +  81  b\ 


QUADRATIC   EQUATIONS  149 

15.  2  x2  —  6  xy  +  3  y2  +  5  x  —  7  y  +  2. 

16.  x2--xy  —  2  y2  —  5  #  +  ?/  +  6. 

17.  a2  +  2ak-15&2-3ac  +  176c-4c2. 

18.  3a2-23ab  +  Ub2  +  a  +  31b-10. 

19.  6z2  +  7a?/  +  2?/2-26a;-162/  +  24. 

20.  10a2  +  ^-24?/2  +  26;c  +  542/-12. 

Solve  the  following  equations : 

21.  or5 +  27  =  0.  29.  (a2-4)(3a2+a;-10)  =  0. 

22.  a4-20^  +  64  =  0.  30.  a;7 -729  a?  =  0. 

23.  tf4  +  2*2  +  9  =  0.  3I  X*_9X2  +  U==0. 


24.   x4  +  4a;2  +  9  =  0. 


32.    9  a*4  — 2a*  +  4  =  0. 


25.  (#+2)(3a2+4a+5)=0. 

26.  *«-64  =  0.  33'  ^-16  =  °« 

27.  2  or* -3  a2 +  4  a- 6  =  0.  34-  2^+6^-18^-54  =  0. 

28.  ^-2^  +  5^  =  0.  35.  (4a?-l)(a?  +  a>+l)  =  0.. 

SIMULTANEOUS   QUADRATIC   EQUATIONS 

223.  In  solving  simultaneous  quadratic  equations  involving 
two  unknown  numbers  it  is  necessary  to  eliminate  one  of  the 
unknowns  as  was  done  in  simultaneous  linear  equations. 

The  elimination  of  an  unknown  number  from  two  equations 
of  the  second  degree  will  often  produce  an  equation  of  the 
fourth  degree  with  one  unknown  number  which  cannot  be 
solved  by  the  ordinary  methods.  The  following  general  direc- 
tions will  lead  to  the  solution  of  many  types. 

224.  Case  I.      When  each  equation  is  in  the  form 

ax2  +  by2  =  c. 
In  this  case,  either  x2  or  y2  can  be  eliminated  by  addition 
or  subtraction  (§  42,  II,  III). 

Case  II.  When  each  equation  is  of  the  second  degree,  and 
homogeneous;  that  is,  when  each  term  involving  the  unknown 
numbers  is  of  the  second  degree  with  respect  to  them  (§  23). 


150  algp:bra 

The  equations  may  then  be  solved  as  follows : 

Ex.    Solve  the  equations     J  '  ^' 

I    x2  +  2/2  =  29.  (2) 

Dividing  (1)  by  (2),  x2~^xV  =  A    0r  29  z2-58  xy  =  5 a?  +  5 i/2. 
a?  +  ?/2      29 

Then,  5  y2  +  58  a;y  -  24  z2  =  0,  or  (5  y  -  2  a:)  (y  +  12  *)  =  0. 

2  a* 
Placing  5  y  —  2  x  =  0,  ^  =  ^ ;  substituting  in  (1), 

5 

z2  _  »£L  _  5?  or  34  =  25. 
5 

Then,  a;  =  ±  5,  and  y  =  —  =  ±  2. 

5 

Case  III.  When  the  given  equations  are  symmetrical  with 
respect  to  x  and  y  ;  that  is,  when  x  and  y  can  be  interchanged 
without  changing  the  equation. 

Equations  of  this  kind  may  be  solved  by  combining  them  in 
such  a  way  as  to  obtain  the  values  of  x  +  y  and  x  —  y. 

\x  +  y  =  2.  (1) 

i.    Solve  the  equations         { 

\        xy=-15.  (2) 

Squaring  (1),  x2  +  2  xy  +  y2  =       4. 

Multiplying  (2)  by  4,  \xy  =—  60. 

Subtracting,  x2  —  2  xy  -f  y2  ==     64. 

Extracting  square  roots,  x  —  y  =  ±  8.  (3) 

Adding  (1)  and  (3),  2  as  =  2  ±  8  =  10  or  -  6. 

Whence,  x  =  5  or  —  3. 

Subtracting  (3)  from  (1),  2  ?/  =  2  T  8  =  -  6  or  10. 

Whence,  y  =  —  3  or  5. 

The  solution  is  x  =  5,    y  =  —  3  ;   or,   x  =  —  3,  y  =  5. 

The  above  method  offers  the  most  desirable  form  of  solution 
and  should  be  employed  when  possible. 

If  one  equation  is  of  the  second  degree,  the  other  of  the 
first  degree,  and  they  are  not  symmetrical,  Case  IV  should  be 
used. 


QUADRATIC    EQUATIONS  151 

Case  IV.  When  one  equation  is  of  the  second  degree  and  the 
other  of  the  first. 

Equations  of  this  kind  may  be  solved  by  finding  one  of  the 
unknown  numbers  in  terms  of  the  other  from  the  first  degree 
equation,  and  substituting  this  value  in  the  other  equation. 

r,      a  i      ^  \2x2-xy  =  6y.  (1) 

Ex.    Solve  the  equations      \  w 

J    x  +  2y  =  l.  (2) 

From  (2),  2  y  5=  7  -  x,  or  y  =  I— £.  (*) 

Substituting  in  (1),  2x2-x(7-^-\  =  qH-^JSV 

Clearing  of  fractions,  4  x2  —  7  x  -f  x2=  42— 6  x,  or  5  x2— x=42. 

Solving,  x  =  3  or  -  — . 

5 

Substituting  in  (3) ,  y  =  1=1?  or  1±J£  =  2  or  —  . 

.....  2  2  10 

The  solution  is   x  =  3,  y  =  2  ;  or  x  =  —  ^,    y  =  fg. 

Certain  examples  where  one  equation  is  of  the  third  degree  and  the 
other  of  the  first  may  be  solved  by  the  method  of  Case  IV. 

225.  Special  Methods  for  the  Solution  of  Simultaneous  Equa- 
tions of  Higher  Degree. 

No  general  rules  can  be  given  for  examples  which  do  not 
come  under  the  cases  just  considered;  various  artifices  are 
employed,  familiarity  with  which  can  only  be  gained  by 
experience. 

v-y=i9.  (i) 


i.    Solve  the  equations 

\x2y-xy2=6.  (2) 

Multiply  (2)  by  3,                           3  x2y  -  3  xy2  =  18.  (3) 
Subtract  (3)  from  (1),  x3  -  Sx2y  +  Sxy2  -  y3  =  1. 

Extracting  cube  roots,                                x  —  y  =  l.  (4) 

Dividing  (2)  by  (4),                                         xy  =  6.  ~               (5) 

Solving  equations  (4)  and  (5)  by  the  method  of  §  224,  Case  III,  we 
find  %  ss  3,  y  =  2  j  or,  x  =  —  2,  y  =—  3. 


152  ALGEBRA 

.  (  x8  -\- y8  =  9  xy. 

2.    Solve  the  equations    1 

^  I    a; +  0  =  6. 

Putting  x  =  u  -f  v  and  y  =  u  —  v, 
(u  +  v)3  +  (w-v)3  =  9(w4-v)(w-'y),  or,  2  w8  +  6  uv2  =  9(u2-  v2);  (1) 
and  (u  -f  a)  +  (m  —  «?)  =  6,  2  w  =  6,  or  w  =  3. 

Putting  w  =  3  in  (1),    54  +  18  v2  =  9  (9  -  t>2). 

Whence,  v2  =  1,  or  v  =  ±  1. 

Therefore,  x  =  w  +  v  =3  ±1=4  or  2  ; 

and  y  =  u  —  v  =  3=Fl=2or4. 

The  solution  is  x  =  4,  ?/  =  2  ;  or,  x  =  2,  y  =  4. 

The  artifice  of  substituting  u  +  v  and  tt  —  v  for  £  and  ?/  is  advantageous 
in  any  case  where  the  given  equations  are  symmetrical  (§  224,  Case  III) 
with  respect  to  x  and  y.     See  also  Ex.  4. 


3.    Solve  the  equations 


x*  +  if  +  2x  +  2y  =  23.  (1) 

xy  =  6.  (2) 

Multiplying  (2)  by  2,  2xy=  12.  (3) 

Add  (1)  and  (3),       x2  +  2xy  +  y2  +  2x +  2y  =  35. 
Or,  (x  +  ?/)2  +  2(x  +  ?/)=35. 

Completing  the  square,  (x  +  y)2  +  2  (x  +  y)  +1  =  36. 
Then,  (x  +  y)  +  1  =  ±  6  ;  and  x  +  y  =  5  or  —  7.  (4) 

Squaring  (4),  x2  +  2  x^  +  y2  =  25  or  49. 

Multiplying  (2)  by  4,  4  xy  =  24. 

Subtracting,  x2  —  2  xy  +  ?/2  =  1  or  25. 

Whence,  x  —  y  =  ±  1  or  ±  5.  (5) 

Adding  (4)  and  (5),  2x  =  5±l,  or-7±6. 

Whence,  x  =  3,  2,  -  1,  or  —  6. 

Subtracting  (5)  from  (4),  2  y  =  5  T  1 ,.  or  -  7  =F  5. 

Whence,  y  =  2,  3,  -  0,  or  -  1. 

The  solution  isx  =  3,  ?/  =  2  ;  x  =  2,  y  =  3  ;  x  =—  1,  ?/  =  —  (J  ;  or  x  =  —  6, 


QUADRATIC    EQUATIONS 


153 


f  xA  +  y*  =  97. 
4.    Solve  the  equations    -j 
4  H  la?  +2/   =-1. 


Putting  x  =  u  +  v  and  y  =  u  —  v, 

(u  +  t>)4  +  (fS  -  *)*  =  97,  or  2  «*  +  12  m¥  +  2  «*  =  97,  (1) 

and  (w  +  a)  +  (w  —  a)  =  —  1,  2  w  =  —  1,  or  w  =  —  J. 

Substituting  value  of  U  in  (1),   J  +  3  v2  +  2  v4  =  97. 


Solving  this, 


25  _      31 


or ;  and  v  =±  -  or  ± 


V-31 

2 


Then,x  =  n+^-l±g,  or -1^^-^1=2,-3,  or~  1±V-:il: 

2      2'  2  2  '  2 


and  y  =  ^-^-lTg      r_l      V-31^_3         r-l^V33T 

2      2  2  2  2 


The  solution  is  x  =  2,    ?/ =  —  3;    x=-3,    ?/  =  2  ;    x- — l  +     — ~, 


_1_V-31     _      -l-V-31    B       -i+V-31 

w  = ;  or  x  = ,  y  = ! 

y  2    .  2  2 


MISCELLANEOUS   EXAMPLES 
EXERCISE   50 

Solve  the  following  equations  and  verify  each  result : 


3- 


(2xy  +  x  =  —  36. 
1  xy  —  3  y  =  —  5. 

6. 

[a2  +  2/2  +  z-2/  =  32. 
1  xy  =  6. 

1-1=12. 

( x2  —  3  xy  —  4:  y2  =  0. 
I3x-5y  =46. 

ja2__2?/2  +  3a:  =  -8. 

a;                 0 

7-     ■ 

8.     i 

^    ?/__io; 

rtfi       3 

U2_2r°-42/  =  -2. 

154 


10. 


i5- 


i8. 


19. 


22. 


23- 


k      ,2     3  b 
x       a 

(x4  +  y4  =  17. 
{  x  —  y  =0. 

(±d  +  k-3dk  =  -6. 
\d-5k  +  2dk  =  10. 

(x2  +  ±xy  =  13. 
[2xy  +  9y2=87.  . 

17. 


ALGEBRA 

r  1 

,  1 

1 

+  — 

+ 

#2 

spy 

.T 

12. 

ji 

,  1 

- 

-h-  = 

s. 

[x 

2/ 

13. 


14. 


16. 


:49. 


a?  +  y  *=»  35. 

Va?  4-  S/y  =  5. 

11  a;2  —  ay  —  y2  =  45. 
7a;2  +  3a^-2?/2  =  20. 

r  ^2  -  24  xy  +  95  =  0. 


13. 


3a?+2y     3a?-2y 


3#— 2  y 
[8y2  +  3a2  =  29, 

-  =  6a2. 
a# 

a;  +  y  =  5  a#y. 

'-  +  ±  =  -19  0.3. 

x6      y6 

-  +  -  =  —  a. 
3      2/ 

e2  +  9Z2  +  4e  =  9. 

ej  4. 2  £  =  -  2. 


3  x2  -  5  ay  =  2  a2  + 1 3  a  b  -  7  b 2 , 
x  +  y  =  3(a  —  6). 

41 

3a,-+2y     20 


.  a;2?/  +  a;?/2 


2a8  +  24a. 
=  2  a3  -  8  a. 


of  4-  yf 
jr*y  +  i 

V2a2-9  =  3y  +  6. 


Va?4  — 17  y2  =  x2  —  5. 


24. 


25. 


26. 


27. 


28. 


29. 


5x-2y  =  l. 

4  a;  +  3  z  =  —  5. 

a?2?/  -}-  #?/2  =  56. 

»  +  y  =  - 1. 

x[  ,  y2  =  19 
y      a       6  * 

a;     y     6 

f3a;2  +  3?/2  =  10a*/. 

[x     y     3 
#2#  -f  y2x  =  42. 

[  X     y     b 

J  5  f  +  ga  -  3  s2  =  27. 

l4^-4gs  +  3s2  =  72. 


QUADRATIC   EQUATIONS  155 

+  4xy-3y  =  42.  .    (  xA  +  tftf  +  yA  =  481. 


'  2  y2  —  xy  +  5  y  =  — 10.  [  a?2  —  #?/  -f  y2  =  37. 

■16  aY- 104  xy=-105.   J         J  9 aj2-13«y-3aj=  -123. 

|a;_2/  =  _2.  33'     1b?/4-42/2  +  22/  =  125. 

*  Divide  the  first  equation  by  the  second. 


EXERCISE   51 

i.  Find  two  numbers  whose  product  is  112  and  whose  dif- 
ference is  6. 

2.  A  rectangular  field  has  a  perimeter  of  104  rods  and  an 
area  of  4  acres.     Find  its  dimensions. 

3.  The  square  of  the  sum  of  two  numbers  minus  four  times 
their  product  equals  49,  and  the  difference  of  their  squares 
equals  175.     What  are  the  numbers  ? 

4.  The  sum  of  the  cubes  of  two  numbers  is  855 ;  and  if  the 
sum  of  the  numbers  be  multiplied  by  their  product,  the  result 
will  be  840.     What  are  the  numbers  ? 

5.  There  is  a  number  consisting  of  two  digits,  the  sum  of 
whose  squares  is  80 ;  and  if  the  sum  of  the  digits  be  multiplied 
by  4,  the  number  will  be  expressed  with  its  digits  reversed. 
What  is  the  number  ? 

6.  A  man  loaned  a  sum  of  money  at  6  %  for  a  given  time 
and  received  $240  interest;  if  he  had  loaned  the  same  sum  for 
two  years  longer  at  the  rate  represented  by  the  first  number 
of  years,  he  would  have  received  $40  more  than  at  first.  Find 
the  time  and  the  amount  loaned. 

7.  If  5  be  added  to  the  denominator  and  subtracted  from 
the  numerator  of  a  certain  fraction,  it  will  be  expressed  by  its 
reciprocal ;  and  the  difference  of  the  squares  of  numerator  and 
denominator  equals  65.     What  is  the  fraction  ? 

8.  A  number  consists  of  three  digits,  the  second  of  which 
is  twice  the  first.     The  sum  of  the  squares  of  the  digits  equals 


156  ALGEBRA 

89,  and  if  99  be  subtracted  from  the  number,  the  digits  will  be 
reversed.     What  is  the  number  ? 

g.  A  man  buys  two  pieces  of  cloth,  each  containing  as  many- 
yards  as  its  price  per  yard  in  cents,  and  he  pays  $  41  for  the 
whole  amount.  If  the  prices  for  the  two  pieces  of  cloth  had 
been  interchanged,  his  bill  would  have  been  $1  less.  How 
many  yards  of  each  did  he  buy  and  what  was  the  price  per 
yard  ? 

io.  Two  squares  have  together  an  area  of  613  square  rods. 
If  the  side  of  the  first  square  were  decreased  by  6,  and  that  of 
the  second  increased  by  1,  their  perimeters  would  be  in  the 
ratio  of  2  to  3.     Find  the  side  of  each  square. 

ii.  There  are  two  numbers  whose  sum  decreased  by  the 
square  root  of  their  product  is  13  ;  and  the  sum  of  their  squares 
increased  by  their  product  is  481.     Find  the  numbers. 

12.  Two  boys  count  their  pennies.  They  find  that  the 
product  of  the  numbers  representing  them  is  84,  and  that  the 
square  of  their  sum  decreased  by  twice  their  difference  is  351. 
How  many  did  each  have  ? 

13.  There  are  two  numbers  whose  difference  is  819,  and 
the  difference  of  their  cube  roots  is  3.  What  are  the 
numbers  ? 

14.  There  is  a  difference  of  one  hour's  time  in  two  trains 
which  go  from  A  to  B,  the  rate  of  the  first  train  being  5  miles 
an  hour  more  than  that  of  the  second  train.  If  the  speed  of 
each  train  were  increased  2  miles  per  hour,  the  difference 
in  time  from  A  to  B  would  be  decreased  7  minutes  80 
seconds.  Find  the  distance  from  A  to  B  and  the  rate  of 
each  train. 

15.  The  difference  of  the  perimeters  of  a  square  and  a  circle 
is  5.752  feet  and  the  circle  contains  81.86  square  feet  more 
than  the  square.  Find  the  radius  of  the  circle  and  the  side 
of  the  square. 


QUADRATIC    EQUATIONS 


157 


16.  In  an  isosceles  triangle  the  product  of  the  base  and  one 
leg  is  108,  and  the  difference  between  the  squares  of  the  base 
and  leg  is  52.     Find  the  altitude  of  the  triangle. 

17.  The  perimeter  of  a  rectangle  is  46  inches.  If  its  length 
be  increased  3  inches,  its  area  will  be  153  square  inches.  Find 
its  dimensions.  Is  there  more  than  one  such  rectangle  ?  Ex- 
plain. 

18.  If  the  sum  of  the  denominator  and  numerator  of  a  cer- 
tain fraction  be  divided  by  their  difference,  the  quotient  is  9. 
But  if  the  product  of  the  numerator  and  denominator  be  di- 
vided by  their  sum,  the  quotient  is  2  with  a  remainder  of  2. 
Find  the  fraction.  What  principle  of  proportion  is  illustrated 
in  this  problem  ?  If  this  principle  is  applied,  are  simultaneous 
equations  necessary  ? 

226.  It  was  noted  in  §§  224,  225,  that  two  second  degree 
equations  had  four  solutions,  or  pairs  of  values  for  x  and  y,  that 
a  second  degree  and  a  first  degree  equation  had  two  solutions, 
that  if  imaginary  roots  entered  they  were  always  in  pairs. 
The  geometric  explanation  for  this  is  readily  seen  if  the 
equations  are  plotted. 

Ex.  1.     Consider  the  equation  x2  -f-  y2  =  25. 

This  means  that,  for  any  point  on  the 
graph,  the  square  of  the  abscissa,  plus  the 
square  of  the  ordinate,  equals  25. 

But  the  square  of  the  abscissa  of  any 
point,  plus  the  square  of  the  ordinate,  equals 
the  square  of  the  distance  of  the  point  from 
the  origin  ;  for  the  distance  is  the  hypotenuse 
of  a  right  triangle,  whose  other  two  sides  are 
the  abscissa  and  ordinate. 

Then  the  square  of  the  distance  from  0  of 
any  point  on  the  graph  is  25  ;  or,  the  distance 
from  0  of  any  point  on  the  graph  is  5. 

Thus,  the  graph  is  a  circle  of  radius  5,  having  its  center  at  0. 

(The  graph  of  any  equation  of  the  form  x2  -f  y2  =  a  is  a  circle.) 


Y 

B 

s 

H 

/ 

\ 

r 

/ 

• 

\ 

X 

/ 

/ 

/ 

1 

/ 

X 

A 

I 

j 

\ 

/ 

\ 

/ 

V 

*• 

^ 

y 

a 

158 


ALGEBRA 


Ex.  2.     Consider  the  equation  y2  =  4  x  +  4 

Ha;-*),       */2  =  4,  or  y  =±2..  (^,  i?) 

If  x  =  1,       y2  =  8,  or  y  =  ±  2  V2.  (C,  Z>) 

If  x=-  1,  y   =  0,     Etc.  (E) 

The  graph  extends  ^definitely  to  the  right  of 
77'. 

If  x  is  negative  and  <  —  1,  y2  is  negative,  and 
therefore  y  imaginary;  then,  no  part  of  the 
graph  lies  to  the  left  of  E. 

(The  graph  of  Ex.  2  is  a  parabola  ;  as  also  is 
the  graph  of  any  equation  of  the  form  y'2  =  ax  or  y2  =  ax  +  b.) 

Ex.  3.     Consider  the  equation  x2  -+-  4  y2  =  4. 

In  this  case  it  is  convenient  to  first 
locate  the  points  where  the  graph  inter- 
sects the  axes. 

If  y  =  0,     x2  =  4,  or  x  =  ±  2.    (.4,  4') 

If  x  =  0,  4  y2  =  4,  or  y  =  ±  1.    (J3,  i?') 

Putting  x  —  ±  1 ,    4  ?/2  =  3,    ?/2  =  J,    or 


V3 


(C,  A  C,  2>') 


| 

Y 

d! 

B 

X 

C 

X 

A' 

? 

v 

Wf 

b' 

D 

, 

y' 

i 

If  x  has  any  value  >  2,  or  <  —  2,  y2  is 
negative,  and  y  imaginary  ;  then,  no  part  of  the  graph  lies  to  the  right  of 
A,  or  left  of  A'. 

If  y  has  any  value  >  1,  or  <  —  1,  x2  is  negative,  and  x  imaginary  ;  then, 
no  part  of  the  graph  lies  above  B,  or  below  B1- 

(The  graph  of  Ex.  3  is  an  ellipse  ;  as  also  is  the  graph  of  any  equation 
of  the  form  ax2  +  by2  —  c. ) 


Ex.  4.     Consider  the  equation  x2  —  2  y2  - 


1. 


Y 

3 

B 

A 

1 

X 

A 

0 

V 

s 

1 

s 

1 

/ 

s 

p 

V 

** 

s 

V 

Y, 

Here  x2  -  1  =  2  y2,  or  y2  =  - 


1 


If  ac  =  ±  1,  2/2  =  0,  ory  =0.     (A'.  A) 

If  x  has  any  value  between  1  and 
—  1,  y2  is  negative,  and  //  imaginary. 

Then,  no  part  of  the  graph  l&Ofl  bt- 
tween  ^4  and  A'. 

If  »:  =±2,  **9t|t  or  y=±y/l. 
(2?,  C,  5',  C) 


QUADRATIC   EQUATIONS 


159 


The  graph  has  two  branches  BAC  and  B'A'C,  each  of  which  extends 
to  an  indefinitely  great  distance  from  O. 

(The  graph  of  Ex.   4  is  a  hyperbola;   as  also  is  the  graph  of  any 
equation  of  the  form  ax2  —  by2  =  c,  or  xy  =  a.) 


Ex.  1.     Consider  the  equations 


227.  Graphical  Representation  of  Solutions  of  Simultaneous 
Quadratic  Equations. 

\y2  =  ±x, 
[3x-y  =  5. 

The  grapjh  of  y2  =  4  x  is  the  parabola  AOB. 

The  graph  of  3  x  —  y  =  5  is  the  straight  line  AB, 
intersecting  the  parabola  at  the  points  A  and  B, 
respectively. 

To  find  the  coordinates  of  A  and  J5,  we  proceed 
as  in  §  48  ;  that  is,  we  solve  the  given  equations. 

The  solution  is  x  =  1,  y  =  —  2  ;  or,  x  =  -^5, 
y  =  ^  (§224,  IV). 

It  may  be  verified  in  the  figure  that  these  are 
the  coordinates  of  A  and  2?,  respectively. 

Hence,  if  any  two  graphs  intersect,  the  coordinates  of  any  point 
of  intersection  form  a  solution  of  the  set  of  equations  represented 
by  the  graphs. 


— Y- j r*C- 

S       i±j ~3. 

III  IstlUl^IZ" 

:z:=Ei^s;::E::: 

— -Y'-t ^s:- 


Ex.  2.     Consider  the  equations 


x2  +  y2  =  17, 
xy  =  4:. 


The  graph  of  x2  -f  y2  =  17  is  the  circle 
AD,  whose  centre  is  at  O,  and  radius  Vl7. 
The  graph  of  xy  =  4  is  a  hyperbola, 
having  its  branches  in  the  angles  XOY 
and  J'07',  respectively,  and  intersecting 
the  circle  at  the  points  A  and  B  in  angle 
XOY,  and  at  the  points  C  and  D  in 
angle  X'OY'. 

The  solution  of  the  given  equation  is 
X  =  4,   ?/  =  1  ;    se  =  1,^  =  4; 
x  —  —  1,  y  =—  4  ;    and  #  =—  4,  y  =  —  1. 
It  may  be  verified  in  the  figure  that  these  are  the  coordinates  of  A,  2?, 
C,  and  D,  respectively. 


V 

\!  1 

*B 

,J 

tt 

r 

*N 

/ 

\ 

\ 

' 

L 

Sk 

1 

\ 

J 

1 

I 

\ 

\ 

«*. 

X 

I 

X 

c 

L 

1 

J 

y 

f 

• 

/ 

' 

X 

' 

-  C1P 

; 

1  1 

v 

! 

160 


ALGEBRA 


2. 


x2  +  4  y2  =  4. 
a  — y  =  l. 

4- 

x2-4y  =  -7. 

2x  +  3y  =  ±. 

5- 

9x2  +  y2  =  U$. 
xy  =  -8. 

6. 

Y 

'""  R 

< 

^ 

X' 

5B""" 

ziss 

Y'N 

EXERCISE  52 

Find  the  graphs  of  the  following  sets  of  equations,  and  in 
each  case  verify  the  principle  of  §  227 : 

'x2  +  y2  =  29. 
xy  =  10. 

2x2  +  5y2  =  53. 
3  x2  -  4  f  =  -  24. 

x2  +  y2  =  13. 
4:X  —  9y  =  6. 

228.     i£c.  1.   Consider  the  equations 

/a2 +  4  ^  =  4,  (1) 

>2x  +  3y  =  -5.  (2) 

The  graph  of  x2  -f  4 1/'2  =  4  is  the  ellipse 

The  graph  of  2x  +  3?/=— 5  is  the 
straight  line  CD. 

If  2/  or  x  is  eliminated  between  these 
two  equations,  we  find  that  the  resulting 
equation  containing  one  unknown  num- 
ber is  such  that  if  all  the  terms  are  transposed  to  one  member,  that  mem- 
ber is  a  trinomial  perfect  square.  Hence,  the  equation  has  equal  roots 
and  the  line  and  curve  are  tangent  at  A  (  218,  II). 

If  in  Ex.  1,  §  228,  the  second  equation  had  been  2x  +  3y  =  — 10(2), 
the  roots  would  have  been  imaginary  and  the  line  would  not  have  met  the 
ellipse. 

REVIEW  EXAMPLES 

EXERCISE  53 

X2 


Scale :  \  inch. 


Vl-«2  + 


i.  Rednce 


Vl-ar2 


■  to  the  form 


1-x2 


2.  Reduce 


i+j 


i  + 


(x  -  a)2 
x-\-  a 
x—a 


QUADRATIC    EQUATIONS  161 

3.  Reduce  0*  +  O2-  (ex-e~x)2  tQ  the  form  / _2_ 

(ex  +  e_x)2  \ex  +  <r 

4.  Reduce  =  to  the  form  = 

5.  Reduce  (x"  +  x    )' "~  4  to  the  form 


2m  2m 

1+-  *  1 


6.  Reduce 


a?  -f-  V#2  —  a2 


— =  to  the  f  onu  -\/    +     • 
x2  —  a2  x^x  —  a 

a2 

7.  Reduce  ^±^±jD±  +  f^>T*  to  unity. 

a;  +  V#2  -\-y2  x  +  V#2  +  y2 


ax 


(a2  __  /p2\f  j[ 

8.  Reduce v  y  to  the  form  —  > 

a  If        a       V     1  Va2-^2 

9.  5  =  Vr+X2.     If  JT=  V2a-?/~^  find  s  =  JS 

2/  v  y 

2 

10.  Reduce  —  —  to  the  form  — , 

x3  1 

11.  Reduce  _ — -  to  the  form  _ 

Xs  —  x4  —  6  25 


1- 


4ar» 

1  -  2  x*\2 


12.  Reduce  V2  a#  —  x2  to  the  form 

a  (x  —  a)2 


162  ALGEBRA 

13.  2  tan  x  +  (tan  xf  —  3  =  0 ;  find  tan  x. 

14.  2  cos  x  H =  3  ;  find  cos  x. 

cos  a; 


15.  — h  2  cot  x  =  -  VI  +  cot2  # ;  find  cot  x. 

cot  a?  2 

16.  Eeduce       <l  +  x)«  •  x^  -  n(l  +  x)^.  *T  to      »*^_. 

(l  +  a;)2"  (l  +  a)»+1 

17.  £  =  -^(12  +  ^-8.     Evaluate  #  when  x  =  15. 

3V3 


4:  XS 

X 


^]-2x-2x^/l-x4 


v  J  VI  —  X*/ 

18.   Eeduce to  the  form 


af 


4ft+-  1 


2+ 


aA        Vl-tf 
2^-1 


19.  Eeduce —  to  the  form  —  . 

2x-  1+2  Ve2  — #  — 1  -y/x2— x— 1 

'a?-2\*     3  (x-2X-  *  /  _2_ 

va;  +  2y  +4U  +  2y     U  +  2V  .  -         a,-2- 

20.  Eeduce  4 — ! — '- : — ■ — i — 1— : — J—  to  the  form  -= — 7- 

2\*  x~^ 


21.  Eeduce 


(x-2 
U  +  2 


to  the  form 


ea  +  e  a 
2a 

2x  _2x 

a(ea-e"~a) 
4 


ARITHMETIC    PROGRESSION 


163 


1      /-V2ry  —  f 
22.   Reduce  y -\ r-J -j- J—  to  the  form  —7/. 


—  r 

„,2 


23.  Reduce  #  — 


24.  Reduce 


/1  +  W* 
V      y2  y  y 


Ap2 


to  3  #  +  2j9  when  y2  =  4p#. 


2/3 


y-« 


■b2x 


a~y 


b4       .        x2     y2      . 
t° 2-^  when  -^  +  t=  =  1. 


IX.     SERIES 


ARITHMETIC   PROGRESSION 


229.  An  Arithmetic  Progression  is  a  series  of  terms  in  which 
each  term,  after  the  first,  is  obtained  by  adding  to  the  preced- 
ing term  a  constant  number  called  the  Common  Difference. 

Thus,  1,  3,  5,  7,  9,  11,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  2. 

Again,  12,  9,  6,  3,  0,  —  3,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  —  3. 

An  Arithmetic  Progression  is  also  called  an  Arithmetic  Series. 

230.  Given  the  first  term,  a,  the  common  difference,  d,  and  the 
number  of  terms,  n,  to  find  the  last  term,  I. 

The  progression  is  a,  a  +  d,  a  -f-  2  d,  a  -j-  3  d,  •••. 
We  observe  that  the  coefficient  of  d  in  any  term  is  less  by  1 
than  the  number  of  the  term. 

Then,  in  the  nth  term  the  coefficient  of  d  will  be  n  —  1. 


That  is, 


1  =  a+  (n  —  l)c?. 


(i) 


164  ALGEBRA 

231.  Given  the  first  term,  a,  the  last  term,  I,  and  the  number 
of  terms,  n,  to  find  the  sum  of  the  terms,  8. 

^  =  a+(a  +  d)+(a  +  2d)  +  ...+(/-d)  +  /. 

Writing  the  terms  in  reverse  order, 

S  =  l  +  (l-d)  +  (l-2d)+  ...  +  (a  +  d)  +  a. 

Adding  these  equations  term  by  term, 

2  5«(a  +  0  +  (a  +  0  +  (a  +  0+»-->(a  +  0  +  («+0- 

Therefore,  2  8  =  n(a  +  T),  and  S  m  £  (a  +  I).  (II) 

m 

232.  Substituting  in  (II)  the  value  of  I  from  (I),  we  have 

£**?[2a+-<»--  l)<q. 

J* 

Ex.   In  the  progression  8,  5,  2,  —  1,   —  4,  •••,  to  27  terms, 
find  the  last  term  and  the  sum. 

Here,  a  =  8,  d  =  5  -  8  m  -  3,  n  =  27. 

Substitute  in  (I),  I  =  8  +(27  -  1)(-  3)  =  8  -  78  =  -  70. 

Substitute  in  (II),  S=  V(8  -  70)  =  27(~  31)  =  -  837- 

The  common  difference  may  be  found  by  subtracting  the  first  term 
from  the  second,  or  any  term  from  the  next  following  term. 

EXERCISE   54 

In  each  of  the  following  find  the  last  term  and  then  the  sum : 

i.  2,  5,  8,  ..-,  to  17  terms. 

2.  3,  9,  15,  «..,  to  12  terms. 

3.  7,  5,  3,  «..,  to  24  terms. 

4.  1,  \,  0,  ..«,  to  32  terms. 

5.  -\,  -T\,  i,  ..-,  to  9  terms. 

6.  a,  a  —  3  b,  a  —  ()b,  •••,  to  Mf  terms. 

7.  2  x  +  5  y,  x  4-  4  y,  3  ?/,  •  •-,  to  13  terms. 


ARITHMETIC   PROGRESSION  165 

0     2c—5dc  —  4:dd  —  c  ■    on  ^,.™0 

8. ,  — - — ,  — — - ,  •••,  to  20  terms. 

3  bo 

3       1  2 

9-     *->  TK~>  —T->  "^  t0  19  terms' 

1         T        P 

I0-     7Vf>   £R>   o£>  "''i  t0  47  terms' 
2o    50    25 

233.  The  j#rs£  term,  common  difference,  number  of  terms,  last 
term,  and  sum  of  the  terms  are  called  the  elements  of  the 
progression. 

If  any  three  of  the  five  elements  of  an  arithmetic  progres- 
sion are  given,  the  other  two  may  be  found  by  substituting  the 
known  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

i .    Given  a  =  —  f,  w  =  20,  S  =  —  f  ;  find  d  and  I. 
Substituting  the  given  values  in  (II), 

-  |  =  10  (-  |  +  I)  or  -  i  =-  |  +1 ;  then,  I  =  J  -  J  =  |. 

Substituting  the  values  of  a,  w,  and  I  in  (I),  f  =—  J  +  19  <Z. 

Whence,  19  J  =  f  +  f  =  ^,  and  d  =  J. 

2.    Given  ^  =  -3,  Z  =  — 39,  £  =  -264;  find  a  and  w. 

Substituting  in  (I),  —  39  =  a  +  (n  -  1)  (-  3),  ora  =  3w-  42.         (1) 
Substituting  the  values  of  Z,  S,  and  a  in  (II), 

-  264  =  -(3  n  -  42  -  39),  or  -  528  =  3  n*  -  81  n,  or  n*  -  27  n  -f  176  =  0. 

wu                               27  ±  V729  -  704      27  ±  5      wtAi,f1 
Whence,  n  =  — == =  — * —  =  16  or  11. 

2  2 

Substituting  in  (1),  a  =  48  —  42  or  33  -  42  =  6  or  —  9. 
The  solution  is  a  =  6,  n  =  16  ;  or,  a  =—  9,  n  =  li. 
The  significance  of  the  two  answers  is  as  follows: 

If  a  =  6  and  n  =  16,  the  progression  is  6,  3,  0,-3,  -  6,  —  9,  —  12, 

-  15,  -  18,  -  21,  -  24,  -  27,  -  30,  -  33,  -  36,  -  39. 
If  a  =  —  9  and  n  =  11 ,  the  progression  is 

_  9,  -  12,  -  15,  -  18,  -  21,  -  24,  -  27,  -  30,  -  33,  -  36,  -  39. 
In  each  of  these  the  sum  is  —  264. 


166  ALGEBRA       ' 

3.    Given  a  =  \,  d  =  —  y1^,  S  =  —  § ;  find  I  and  n. 

Substituting  in  (I),  I  =  |+  (n  -  1)Y  J  -L\  =s  ^L«.  (1) 

Substituting  the  values  of  a,  #,  and  I  in  (II), 

2  2\3     12  y  v  12  y 


,,rl                                      9±V81+144      9  ±15      10  o 

Whence,  71=— ± -J- =  — ± —  =  12  or  —  3. 

■  2  2 

The  value  w  =  —  3  must  be  rejected,  for  the  number  of  terms  in  a  pro- 
gression must  be  a  positive  integer. 

Substituting  n  =  12  in  (1),  I  =  6rf^==-^X. 

A  negative  or  fractional  value  of  w  must  be  rejected,  together  with  all 
other  values  dependent  on  it. 

EXERCISE   55 

i.    Given  a  =*5,  d  =  2,  I  =  65,-,  find  n  and  S. 

2.  Given  cZ  =  —  3,  n  =  42,  I  =  — 119 ;  find  a  and  £. 

3.  Given  d  =  f ,  ri  =  16,  #  =  ^^;  find  a  and  Z. 

4.  Given  71  =  19,  Z  =""""*    ,  S  = ;  find  a  and  d. 

7  14 

5.  Given  S  =  -  540,  a  =  -  23,  n  =  48 ;  find  Z  and  d. 

6.  Given  d  =  —  4, Z  =  —  — — ,  #  =  — ;  find  a  and  n. 

64     '  32       ' 

7.  Given  d  —  a  —  1,  a  =  2  a  +  5,  #  =  44  a  -f  12  ;  find  7  and  n. 

8.  Given  a  =  —  8  a,  I  =  8  a  —  16  Z>,  #  =  —  136  6 ;  find  n  and  d. 

9.  Given  a  =  .4,  I  =  34.6,  n  =  20 ;  find  cZ  and  5. 

10.  Given  S  =  18.15,  d  =  . 02,  a  =  .23;  find  Z  and  n. 

1 1 .  Given  S  =  ^— — ,  d  =  — -^ ,  w  =  15 ;  find  a  and  I. 

12.  Given  n  =  26,  d  =    ,  Z  =  ~         ;  find  #  and  a. 

8  8      ' 


ARITHMETIC   PROGRESSION  167 

234.    From  (I)  and  (II),  general  formulae  for  the  solution  of 
examples  like  the  above  may  be  readily  derived. 

Ex.    Given  a,  d,  and  S\  derive  the  formula  for  n. 

By  §232,  2^=«[2a  +  (»-l)d],  or  dn2  +  (2a-d)n  =  2S. 
This  is  a  quadratic  in  w,  and  may  be  solved  by  the  method  of  §  213; 
multiplying  by  4  d,  and  adding  (2  a  —  d)2  to  both  members, 

4  <Pn2  +  4  d(2  a-d)n  +  (2a-d)2=:$dS  +  (2a-  d)2. 

Extracting  square  roots,    2dn  +  2  a  —  d=±  V8  dS  +  (2  a  —  d)2. 


Whence,  n  =  d  -  2  a  ±  VSdS+^J^2 . 

2d 

EXERCISE   56 

i.  Given  a,  I,  and  n\  derive  the  formula  for  d. 

2.  Given  a,  n,  and  S ;  derive  the  formulae  for  d  and  I. 

3.  Given  d,  n,  and  S ;  derive  the  formulae  for  a  and  Z. 

4.  Given  a,  d,  and  Z ;  derive  the  formulae  for  n  and  S. 

5.  Given  d,  I,  and  71 ;  derive  the  formulae  for  a  and  S. 

6.  Given  Z,  n,  and  $ ;  derive  the  formulae  for  a  and  d. 

7.  Given  a,  cZ,  and  S ;  derive  the  formulae  for  Z. 

8.  Given  a,  Z,  and  # ;  derive  the  formulae  for  d  and  n. 

9.  Given  cZ,  Z,  and  #  ;  derive  the  formulae  for  a  and  n. 

235.    Arithmetic  Means. 

We  define  inserting  m  arithmetic  means  between  two  given 
numbers,  a  and  b,  as  finding  an  arithmetic  progression  of 
?/i  +  2  terms,  whose  first  and  last  terms  are  a  and  b. 

Ex.     Insert  5  arithmetic  means  between  3  and  —  5. 

We  find  an  arithmetic  progression  of   7  terms,  in  which  a  =  3,  and 
I  =  —  5  ;  substituting  n  =  7,  a  =  3,  and  Z  =  —  5  in  (I), 
-  5  =  3  +  6  d,  or  d  =  -  § . 
The  progression  is  3,  ) ,  J,  —  1,  —  J,  —  y,  —  5. 


168  ALGEBRA 

236.  Let  x  denote  the  arithmetical  mean  between  a  and  b. 
Then,  x—  a  =  b  —  x,  or  2  x  —  a -f-  b. 

Whence,  x  =  ^-~-  - 

That-  is,  the  arithmetic  mean  between  two  numbers  equals  one- 
half  their  sum. 

EXERCISE   57 

i.  Insert  6  arithmetic  means  between  3  and  24. 

2.  Insert  12  arithmetic  means  between  —  5  and  73. 

3.  Insert  20  arithmetic  means  between  |-  and  —  f-f. 

4.  Insert  13  arithmetic  means  between  —  -*-  and  —  -2g3-. 

5.  Find  the  arithmetic  mean  between  a2  —  2  a  —  9  and 
a2_6a  +  l. 

6.  If  n  —  2  arithmetic  means  are  inserted  between  a  and  I, 
find  the  4th  term. 

GEOMETRIC   PROGRESSION 

237.  A  Geometric  Progression  is  a  series  of  terms  in  which 
each  term,  after  the  first,  is  obtained  by  multiplying  the  pre- 
ceding term  by  a  constant  number  called  the  Ratio. 

Thus,  2,  6,  18,  54,  162,  •••  is  a  geometric  progression  in 
which  the  ratio  is  3. 

9,  3, 1,  i,  |-,  •  •  •  is  a  geometric  progression  in  which  the  ratio  is  \. 

—  3,  6,  — 12,  24,  —  48,  •  •  •  is  a  geometric  progression  in 
which  the  ratio  is  —  2. 

A  Geometric  Progression  is  also  called  a  Geometric  Series. 

238.  Given  the  JJrsf  term,  <i,  the  ratio,  r,  and  the  number  of 
terms,  n,  to  find  th€  lax/  term,  I. 

The  progression  is  a,  ar,  ar2,  a?*3,  •••. 

We  observe  that  the  exponent  of  r  in  any  term  is  less  by  1 
than  the  number  of  the  term. 

Then,  in  the  nth  term  the  exponent  of  r  will  be  n  —  1. 

That  is,  l  =  arn-K  (I) 


GEOMETR&    PROGRESSION  169 

239.  Given  the  first  term,  a,  the  last  term,  I,  and  the  ratio,  r,  to 
find  the  sum  of  the  terms,  8. 

S  =  a-\-ar  +  ar2-\-  •••  +  arn~*  -\-arn-2 +  ai"-\  (1) 

Multiplying  each  term  by  r, 

rjS  =  ar  =  ar2  +  ar*+  ...  +  ar1l~2  +  ar"-1  +  arn.  (2) 

Subtracting  (1)  from  (2),  rS  —  S  =  arn  —  a,  or  S  = ~~i~' 

r  —  1 

But  by  (I),  §  238,  rl=arn. 

Therefore,  JS  = 7^^  -  (II) 

r  — 1 

The^rs^  Jerm,  ratio,  number  of  terms,  last  term,  and  sum  of  the  terms 
are  called  the  elements  of  the  progression. 

240.  Examples. 

i.   In  the  progression  3,  1,  \,  •••,  to  7  terms,  find  the  last 
term  and  the  sum. 

Here,  0  =  3,  r  =  \,  n  =  7. 

Substituting  in  (I),  l  =  s(-Y  =  ~  =  — . 

°       V  *  \S)       35      243 


Substituting  in  (II),  S  - 


i-1  -1  -I         ^3 


The  ratio  may  be  found  by  dividing  the  second  term  by  the  first,  or 
any  term  by  the  next  preceding  term. 

2.    In  the  progression  —  2,  6,  —18,  •••,  to  8  terms,  find  the 
last  term  and  the  sum. 

Here,  a  =  —  2,  r  = =  —  3,  n  =  8,  therefore, 

I  =  -  2(-  3)'  =  -2x(-  2187)  =  4374. 

#  -  -  3  x  4374  -  (-  2)  _  -  13122  +  2  =  328Q 
—  3  —  X  —  4 


170  ALGEBRA 

EXERCISE   58 

Find  the  last  term  and  sum  of  the  following : 
i.    1,  3,  9,  •••  to  8  terms. 


2. 

2,  1,  i,  •••  to  11  terms. 

3- 

5,  -10,  20,  •••  to  12  terms. 

4- 

-iTir>  -&*  '"  to  7  terms. 

5- 

h  h  h  *  •  •  to  6  terms. 

6. 

-  h  -tV>  —  Hi  ••'  t0  8  terms. 

241.  If  any  three  of  the  five  elements  of  a  geometric  pro- 
gression are  given,  the  other  two  may  be  found  by  substituting 
the  given  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

But  in  certain  cases  the  operation  involves  the  solution  of 
an  equation  of  a  degree  higher  than  the  second ;  and  in  others 
the  unknown  number  appears  as  an  exponent,  the  solution  of 
which  form  of  equation  can  usually  only  be  effected  by  the  aid 
of  logarithms  (§  110). 

i.    Given  a  =  —  2,  n  =  5,  I  =  —  32 ;  find  r  and  S. 
Substituting  the  given  values  in  (I),  we  have 

—  32  =  —  2  r4 ;  whence,  r4  =  16,  or  r  =  ±  2. 

Substituting  in  (II), 

If  r  =  2,       S  =  2(-32)-(-2)  =  _  64      2  _  _  62 

2-1 

If  r=_o    ig=(-2)(-32)-(-2)  =  64  +  2=^22 

-2-1  -3 

The  solution  is  r  =  2,  S  =  -  62  ;  or,  r  =  -  2,  S  =-  22. 
The  interpretation  of  the  two  answers  is  as  follows : 
If  r  =  2,  the  progression  is  —  2,  —  4,  —  8,  —  16,  —  32,  whose  sum  is 
-62. 

If  r  =—  2,  the  progression  is  —  2,  4,  —  8,  16,  —  32,  whose  sum  is  —  22. 


GEOMETRIC   PROGRESSION  171 

2.    Given  a  =  3,  r  =  -  £,  #  =  VsV" ;  find  w  and  Z. 

Substituting  in  (II),     ^  =  -**~8  =  L±i. 

°       l    ;'      729        -}-l  4 

Whence,  1  +  9  =  tfiff  -t  or,  I  =  V#  -  9  =-  7fc. 

Substituting  the  values  of  a,  r,  and  I  in  (I), 

-  th  =  H- i)""1 ;  or,  (-  J)"-1  =- nVy. 

Whence,  by  inspection,  n  —  1=7,  orn  =  8. 

From  (I)  and  (II)  general  formulae  may  be  derived  for  the  solution  of 
cases  like  the  above. 

If  the  given  elements  are  n,  Z,  and  S,  equations  for  a  and  r  may  be 
found,  but  there  are  no  definite  formulas  for  their  values. 

The  same  is  the  case  when  the  given  elements  are  a,  n,  and  S. 

The  general  formulae  for  n  involve  logarithms ;  these  cases  are  discussed 
in  §  110. 

EXERCISE   59 

i.  Given  r  =  2,  n  =  12,  S  =  4095  ;  find  a  and  I 

2.  Given  a  =  2,  r  =  —  3,  Z  =  1458 ;  find  n  and  S. 

3.  Given  Z  =  —  -g^,  a  =  —  if ,  n  =  10 ;  find  r  and  S. 

4.  Given  a  »  f,  J  a  3584,  #  =  **£** ;  find  r  and  rc. 

5.  Given  r  =  i,  w  =  5,  Z  =  T^ ;  find  a  and  S. 

6.  Given  ^  =  -A||-i^  a  =  -64,  r=i;  find  w  and  I 

7.  Given  a,  Z,  and  # ;  derive  the  formula  for  r. 

8.  Given  r,  Z,  and  n ;  derive  the  formulae  for  a  and  S. 

9.  Given  a,  n,  and  Z ;  derive  the  formulae  for  r  and  S. 
10.  Given  #,  n,  and  r ;  derive  the  formulae  for  a  and  L 

242.    Sum  of  a  Geometric  Progression  to  Infinity. 

The  limit  (§  125)  to  which  the  sum  of  the  terms  of  a  decreas- 
ing geometric  progression    approaches,    when    the    number  of 


172  ALGEBRA 

terms  is  indefinitely  increased,  is  called  the  sum  of  the  series  to 
infinity. 

Formula  (II),  §  239,  may  be  written 

0      a—  rl 
1  —  r 

It  is  evident  that,  by  sufficiently  continuing  a  decreasing 
geometric  progression,  the  absolute  value  of  the  last  term  may 
be  made  less  than  any  assigned  number,  however  small. 

Hence,  when  the  number  of  terms  is  indefinitely  increased, 
Z,  and  therefore  rl,  approaches  the  limit  0. 

Then,  the  fraction  a~~r   approaches  the  limit  — ^—  • 


Therefore,  the  sum  of  a  decreasing  geometric  progression  to 
infinity  is  given  by  the  formula 

Sfatr^z*  (HI) 

1  —  r 
Ex.   Find  the  sum  of  the  series  4,  —  f,  -1/,  ...  to  infinity. 

o 

Here  a  =  4,  r  = 

3 

Substituting  in  (III),  S=     4    =  — . 
V      J  1+f      5 

To  find  the  value  of  a  repeating  decimal. 

This  is  a  case  of  finding  the  sum  of  a  decreasing  geometric 
series  to  infinity,  and  may  be  solved  by  formula  (III). 

Ex.   Findthe  value  of  .85151-... 

We  have,  .85151  •••  =  .8  +  .051  -f  .00051  +  .... 

The  terms  after  the  first  constitute  a  decreasing  geometric  progression 
in  which  a  =  .051,  and  r  —  .01. 

Substituting  in  (III) ,  S  =     ,051     =  '—  =  —  =  —  • 
1-.01       .99       990      330 

Then  the  value  of  the  given  decimal  is  r8^  +  ^,  or  |f  J. 


GEOMETRIC    PROGRESSION  173 

EXERCISE   60 

Find  the  sum  to  infinity  of  the  following : 

T       9     2      2      . . .  4        _  8      _  1  R      _  3  2     . 

2.    1,  -i,i,-.  5-    -.3,  .12,  -.048,  -.. 

3-    fj  I?  TV*  '"'  *i  ~~  **  &  ***• 

Find  the  values  of  the  following: 

7-    .4777.-..     8.   .8181.-..     9-   .5243243-...     io.   .207575-... 

243.  Geometric  Means. 

We  define  inserting  m  geometric  means  between  two  numbers, 
a  and  b,  as  finding  a  geometric  progression  of  m  +  2  terms, 
whose  first  and  last  terms  are  a  and  b. 

Ex.    Insert  5  geometric  means  between  2  and  ^ff. 
We   find   a  geometric   progression  of   7   terms,  in  which  a  =  2,  and 
I  =  iff ;  substituting  n  =  7,  a  =  2,  and  Z  =  }||  in  (I), 

i|8  a*  ^  r* ;  whence  r6  =  7%%,  and  r  =  ±  f 
The  result  is  2,  ±  f,  |,  ±  if,  |f,  ±  ^,  fff. 

244.  Let  x  denote  the  geometric  between  a  and  b. 

Then.  -  =  -,  or  x2  =  ab. 

a     x 

Whence,  x  =  ^Jab. 

That  is,  the  geometric  mean  between  two  7i'umbers  is  equal  to 
the  square  root  of  their  product. 

245.  Problems. 

i.  The  sixth  term  of  an  arithmetic  progression  is  f,  and  the 
fifteenth  term  is  -L6-.     Find  the  first  term. 

Ry  §  230,  the  sixth  term  is  a  +  5  d,  and  the  fifteenth  term  a  -f  14  d . 

,       ,  «'".  f  a+    5d  =  J.  .  (1) 

Then,  by  the  conditions,      \  ■ 

la  +  Ud=\*-.  r  (2) 

Subtracting  (1)  from  (2),  9  d  =  }j  whence,  d  =  \. 

Substituting  in  (1),  a  +  §  =  |j  whence,  a  =  -f. 


174  ALGEBRA 

2.  Find  four  numbers  in  arithmetic  progression  such  that 
the  product  of  the  first  and  fourth  shall  be  45,  and  the  product 
of  the  second  and  third  77. 

Let  the  numbers  be  x  —  3  y,  x  —  y,  x  +  y,  and  x  -f  3  y. 
x2-9y2  =  45. 


Then  by  the  conditions,    I   x<*      9  y2  ~  45, 

I   x2-     w2  =  77. 


Solving  these  equations,  x  =  9,  y  —  ±  2  ;  or,  x  =  —  9,  y  =  ±  2  (§  224) 

Then  the  numbers  are  3,  7,  11,  16 ;  or,  —3,  —  7,  —  11,  —  15.   • 

In  problems  like  the  above,  it  is  convenient  to  represent  the  unknown 

numbers  by  symmetrical  expressions. 

Thus,  if  five  numbers  had  been  required,  we  should  have  represented 

them  by  x  —  2  ?/,  x  —  y,  x,  x  +  y,  and  x  +  2  y. 

3.  Find  3  numbers  in  geometric  progression  such  that  their 
sum  shall  be  14,  and  the  sum  of  their  squares  84. 
Let  the  numbers  be  represented  by  a,  ar,  and  af2. 

a  +  qr  +  ar2  =  14.  (1) 

a2  +  a2r2+a2ri  =  te.  (2) 

Divide  (2)  by  (1),  a  -  ar  +  ar2  =  6.  ,  (3) 


Then,  by  the  conditions, 


2 

ar 

=  8,  or  r 

_4 

a 

(4) 

a 

a 

=  14 

,  or  a2- 

-10  a 

+  16 

=  0. 

a  ■■ 

=  8  or  2. 

r 

_4 

8 

or4-  = 
2 

lor 

2 

2. 

Subtract  (3)  from  (1), 

Substituting  in  (1), 

Solving  this  equation, 

Substituting  in  (4) , 

Then,  the  members  are  2,  4,  and  8. 

EXERCISE  61 

1.  The  seventh  term  of  an  A.  P.  is  ■££,  the  twenty-first  term 
is  -3^$.     Find  the  fifteenth  term. 

2.  Show  that  the  sum  of  the  odd  integers  from  1  to  999  is 
the  square  of  their  number. 

3.  The  first  term  of  an  A.  P.  is  1,  the  sum  of  the  third  and 
ninth  terms  is  32.     Find  the  sum  of  the  first  thirteen  terms. 


GEOMETRIC   PROGRESSION  175 

4.  The  sum  of  the  first  ten  terms  of  an  A.  P.  is  to  the  sum 
of  the  first  seven  terms  as  29  to  14.  Find  the  ratio  of  the 
common  difference  to  the  first  term. 

5.  There  are  four  numbers,  such  that  the  first  three  form  a 
G.  P.,  the  last  thr^ee  form  an  A.  P.  The  sum  of  the  first  three 
is  73,  of  the  last  three  192.  The  difference  between  the  second 
and  fourth  is  112.     Find  the  numbers. 

6.  How  many  arithmetic  means  are  inserted  between  —  f 
and  f ,  when  their  sum  is  2^-  ? 

7.  Find  four  numbers  in  A.  P.,  such  that  the  sum  of  the 
first  and  second  shall  be  —  1,  and  the  product  of  the  second 
and  fourth  24. 

8.  A  traveller  sets  out  from  a  certain  place,  and  goes  7 
miles  the  first  hour,  1\  the  second  hour,  8  the  third  hour,  and 
so  on.  After  he  has  been  gone  5  hours,  another  sets  out  and 
travels  16 J  miles  an  hour.  How  many  hours  after  the  first 
starts  are  the  travellers  together  ? 

9.  If  a  person  saves  $  120  each  year,  and  puts  the  sum  at 
simple  interest  at  3\°/o  at  the  end  of  each  year,  to  how  much 
will  his  property  amount  at  the  end  of  18  years  ? 

10.  A  ball  is  dropped  from  a  window  32  feet  above  the 
pavement.  Assuming  the  ball  to  be  perfectly  elastic  and  that 
on  each  rebound  it  rises  to  within  \  of  its  former  height,  how 
far  does  it  travel  before  coming  to  rest  ? 

11.  Two  men  travel  from  P  to  Q,  leaving  P  at  the  same 
time.  The  distance  from  P  to  Q  is  63  miles.  The  first 
travels  1  mile  the  first  hour,  2  miles  the  second  hour,  4  miles 
the  third  hour,  and  so  on.  The  second  travels  11  miles  the 
first  hour,  10|  miles  the  second  hour,  10 1  miles  the  third 
hour,  and  so  on.     Which  is  first  to  arrive  at  Q  ? 

12.  Find  the  geometric  mean  between  .0729  and  .0529. 

13.  Find  the  geometric  mean  between  — and 

2.2o  o76. 


176  ALGEBRA 


14.  Find  the  geometric  mean  between  — TjW  and  ?L. 

xy-y2  xy 

15.  Find  the  geometric  mean  between  a2—  4 a +  4  and 
4  a2  +  4  a  + 1. 

16.  The  product  of  the  first  five  terms  of  a  G.  P.  is  243. 

Find  the  third  term. 

17.  The  digits  of  a  number  of  three  figures  are  in  geometric 
progression.  If  units'  and  tens'  digits  are  interchanged,  the 
number  formed  exceeds  the  original  number  by  36.  The  sum 
of  the  digits  is  14.     Find  the  number. 

18.  A  man  travels  445*-  miles.  He  travels  10  miles  the  first 
day,  and  increases  his  speed  one-half  mile  in  each  succeeding 
day.     How  many  days  does  the  journey  require  ? 

19.  An  A.  P.  has  19  terms  such  that  the  sum  of  the  three 
middle  terms  is  3,  and  the  sum  of  the  first  term  and  the  last 
two  terms  is  —  13.     Find  the  series. 

20.  Find  the  number  of  arithmetic  means  between  1  and  69, 
such  that  the  ratio  of  the  last  mean  to  the  first  mean  is  13. 

21.  Find  an  A.  P.  of  17  terms  such  that  the  sum  of  the  first 
three  terms  is  to  the  last  term  as  3  to  13,  the  first  term  being 
unity. 

22.  The  sum  of  three  successive  terms  of  a  geometric  pro- 
gression is  39  and  the  sum  of  their  squares  is  819.  Find  the 
series. 

23.  The  sum  of  how  many  terms  of  the  series  1,  3,  9  •••,  is 

3280  ? 

24.  Show  that  in  any  G.  P.,  if  each  term  is  subtracted  from 
the  succeeding  term,  the  differences  form  a  G.  P. 

25.  Find  three  numbers  in  A.  P.,  such  that  the  square  of  the 
first  added  to  the  product  of  the  other  two  gives  16,  and  the 
square  of  the  second  added  to  the  product  of  the  other  two 
gives  14. 


PAET   II 


X.   INFINITE    SERIES 


246.   Infinite  Series  (§  178)  may  be  developed  by  Division, 
or  by  Evolution. 

Let  it  be  required,  for  example,  to  divide  1  by  1  —  x. 


1  —  se)l(l  +  x  +  x2+  ■ 
1-x 

X 

x-x2 


.=  l+x  +  x2  +  x?  +  -•. 


Then, 


Again,  let  it  be  required  to  find  the  square  root  of  1  +  #• 


(1) 


1+x 

1  +  ?_£l  +  ... 
2       8 

1 

2  +  ^ 
2 

X 
X    - 

r2 

2  +  x- 

x*\ 
8| 

X2 

4 

Then,    vTTx  =  l +£-!-+•• 


(2) 


It  should  be  observed  that  the  series,  in  (1)  and  (2),  do  not  give  the 
values  of  the  first  members  for  every  value  of  x ;  thus,  if  x  is  a  very 
large  number,  they  evidently  do  not  do  so. 


EXERCISE   62 

Expand  each  of  the  following  to  four  terms : 
T     5-f-4a  l+3# 


2. 


l  +  3aj 

1  +  3  a  +  x2* 


3- 
4. 


5x 
5  —  X  —  3X2 


179 


180  ALGEBRA 


5.    Vl  +  3a.  8.    Va3  +  bs. 


6.    VI  -5x2.  9-    V^  +  l. 


7.    Va2  +  62.  io.    V9a2-16  62. 

CONVERGENOY   AND   DIVERGENCY   OP   SERIES 

247.  An  infinite  series  is  said  to  be  Convergent  when  the 
sum  of  the  first  n  terms  approaches  a  fixed  finite  number  as  a 
limit  (§  125),  when  n  is  indefinitely  increased. 

An  infinite  series  is  said  to  be  Divergent  when  the  sum  of 
the  first  n  terms  can  be  made  numerically  greater  than  any 
assigned  number,  however  great,  by  taking  n  sufficiently  great. 

248.  Consider,  for  example,  the  infinite  series 

1  -{- x  +  x2 -{- Xs  +  •••. 

I.  Suppose  x  —  a?!,  where  xx  is  numerically  <  1. 
The  sum  of  the  first  n  terms  is  now 

1  +  Xl  +  Xl>+  ...  +  Xl»-i  =  l^£Ln(§  103,  VII). 

1  —  Xi 

If  n  be  indefinitely  increased,  x"  decreases  indefinitely  in 
absolute  value,  and  approaches  the  limit  0. 

Then  the  fraction approaches  the  limit . 

1  —  xx  1  —  x1 

That  is,  the  sum  of  the  first  n  terms  approaches  a  fixed  finite 
number  as  a  limit,  when  n  is  indefinitely  increased. 

Hence,  the  series  is  convergent  when  x  is  numerically  <  1. 

II.  Suppose  05  =  1. 

In  this  case,  each  term  of  the  series  is  equal  to  1,  and  the 
sum  of  the  first  n  terms  is  equal  to  n ;  and  this  sum  can  be 
made  to  exceed  any  assigned  number,  however  great,  by  taking 
n  sufficiently  great. 

Hence,  the  series  is  divergent  when  x=l. 


INFINITE   SERIES  181 

III.  Suppose  x  m  —  1. 

In  this  case,  the  series  takes  the  form  1  —  1  +  1  —  1  +  ..., 
and  the  sum  of  the  first  n  terms  is  either  1  or  0  according  as  n 
is  odd  or  even. 

Hence,  the  series  is  neither  convergent  nor  divergent  when 

08-1. 

An  infinite  series  which  is  neither  convergent  nor  divergent 
is  called  an  Oscillating  Series. 

IV.  Suppose  x  =  xly  where  x  is  numerically  >  1. 
The  sum  of  the  first  n  terms  is  now 

1  +  Xl  +  Xl2  +  ...  +  Xin-l  =  ^LJZI  (§    103 ,   VII). 

xx  —  1 

xn  —  1 
By  taking  n  sufficiently  great,  — can  be  made  to  numer- 
al —  1 

ically  exceed  any  assigned  number,  however  great. 

Hence,  the  series  is  divergent  when  x  is  numerically  >  1. 

249.   Consider  the  infinite  series 

l+x  +  x2  +  xs-\ , 

developed  by  the  fraction ■  (§  246). 

1  —  x 

Let  x=  .1,  in  which  case  the  series  is  convergent  (§  248). 
The  series  now  takes  the  form  1  +  .1  +.01  4-  .001  -f-  ••• ,  while 

the  value  of  the  fraction  is  —  or  —  . 

.9        9 

In  this  case,  however  great  the  number  of  terms  taken,  their 
sum  will  never  exactly  equal  y*-. 

But  the  sum  approaches  this  value  as  a  limit;  for  the  series 
is  a  decreasing  geometric  progression,  whose  first  term  is  1,  and 

ratio  .1 ;    and,  by  §  242,  its  sum  to  infinity  is ,  or  — . 

Thus,  if  an  infinite  series  is  convergent,  the  greater  the  num- 
ber of  terms  taken,  the  more  nearly  does  their  sum  approach 


182  ALGEBRA 

the    value    of   the    expression   from    which    the    series    was 
developed. 

Again,  let  x  =.10,  in  which  case  the  series  is  divergent. 

The   series  now   takes  the  form' 1 +10 +  100 +  1000 H , 

while  the  value  of  the  fraction  is — ,  or  —  -. 

In  this  case  the  greater  the  number  of  terms  taken,  the 
more  does  their  sum  diverge  from  the  value  —  J. 

Thus,  if  an  infinite  series  is  divergent,  the  greater  the  num- 
ber of  terms  taken,  the  more  does  their  sum  diverge  from  the 
value  of  the  expression  from  which  the  series  was  developed. 

It  follows  from  the  above  that  an  infinite  series  cannot  be 
used  for  the  purposes  of  demonstration  if  it  is  divergent. 

SUMMATION   OF  SERIES 

250.  The  Summation  of  an  infinite  literal  series  is  the  pro- 
cess of  finding  an  expression  from  which  the  series  may  be 
developed. 

RECURRING   SERIES 

251.  Consider  the  infinite  series 

l  +  2oj  +  3a2  +  4ar3  +  5a4+.... 
Here  (3  x2)  -  2  x(2  x)  +  af  (1)  =  0, 

(4  Xs)  -  2  x(S  x2)  +  x\2  x)  =  0,  etc. 

That  is,  any  three  consecutive  terms',  as,  for  example,  2  x,  3  x2, 
and  4:  Xs,  are  so  related  that  the  third,  minus  2x  times  the 
second,  plus  x2  times  the  first,  equals  0. 

252.  A  Recurring  Series  is  an  infinite  series  of  the  form 

Oo  +  a^  +  o^H — , 
where  any  r  + 1  consecutive  terms,  as  for  example 
anx  ,  au_iX     ,  an_2x     ,  •••,  an_rx     , 


INFINITE   SERIES  188 

are  so  related  that 

anxn  +  px  (a^x"-'1)  +  qx2  (an_^xn-2)  +  ...  +  sxr(an_rxn~r)  =  0 ; 

p,  q}  •  ••,  s  being  constants. 

The  above  recurring  series  is  said  to  be  of  the  rth  order,  and 
the  expression 

1  +  px  -{-px2  -f  •  •  •  +  sxr 

is  called  its  scale  of  relation. 

The  recurring  series  of  §  251  is  of  the  second  order,  and  its 
scale  of  relation  is  1  —  2  x  +  x2. 

An  infinite  geometric  series  is  a  recurring  series  of  the  first  order. 
Thus,  in  the  infinite  geometric  series 

1  +  x+x2+x*  +  .--, 

any  two  consecutive  terms,  as  for  example  x%  and  ic2,  are  so  related  that 
(x3)  —  x(x'2)  =  0  ;  and  the  scale  of  relation  is  1  —  x. 

253.    To  find  the  scale  of  relation  of  a  recurring  series. 

If  the  series  is  cf  the  first  order,  the  scale  of  relation  may  be 
found  by  dividing  any  term  by  the  preceding  term,  and  sub- 
tracting the  result  from  1. 

If  it  is  of  the  second  order,  a0,  a19  a2,  a3,  •••,  its  consecutive 
coefficients,  and  1+px  +  qx2  its  scale  of  relation,  we  shall  have 

'<h+P<ii  +  q<Xo  =  0,  (1) 

^a3+pa2  +  9«i  =  0; 

from  which  p  and  q  may  be  determined. 

If  the  series  is  of  the  third  order,  cr0,  a1?  a2,  a3,  a4,  a5,  •••,  its 
consecutive  coefficients,  and  1  -\-px  -f  qx2  +  rx3  its  scale  of  rela- 
tion, we  shall  have 

'  o-8  +  Vai  +  Qai  +  rao  =  0, 
«4  +i>«3  +  qa2  +  ra^  =  0, 
a5  +pa4  +  qa3  +  ra2  =  0 ; 

from  which  p,  q,  and  r  may  be  determined. 

To  ascertain  the  order  of  a  series,  we  may  first  make  trial  of 
a  scale  of  relation  of  three  terms. 


184  ALGEBRA 

If  the  result  does  not  agree  with  the  series,  try  a  scale  of 
four  terms,  five  terms,  and  so  on  until  the  correct  scale  of  rela- 
tion is  found. 

If  the  series  is  assumed  to  be  of  too  high  an  order,  the  equa- 
tions corresponding  to  the  assumed  scale  will  not  be  inde- 
pendent (§  43). 

254.  To  find  the  sum  (§  250)  of  a  recurring  series  when  its 
scale  of  relation  is  known. 

Let  1  +  px-\-qx2  be  the  scale  of  relation  of  the  series 

a0  +  axx  +  a2x2  -f-  •••. 

Denoting  the  sum  of  the  first  n  terms  by  &ni  we  have 

Sn  =  a0  +  ctix  4  a2x2  + \-  an-ixn-\ 

Then,  pxSn  =  pa^x  +  pa\x2  +  •  •  •  +  pan-2Xn~l  +  pan-\Xn, 

and  qx2Sn  =  qa0x2  +  •••  -h  qan-sx11-1  -f  qan-2Xn  +  qan-\Xn+l. 

Adding  these  equations,  and  remembering  that,  by  virtue  of  the  scale 
of  relation, 

«2  +pai  4-  qa0  =  0,    •  •,  an-i  +pan-2  +  qan-z  =  0, 

the  coefficients  of  x2,  xs,  •  ••,  a:n_1  become  0,  and  we  have 

8n0.  +px  +  qx2)  =  a0  +  («i  +pa0)x  4-  (pan-i  4-  qan-2)xn  4-  ^n-i«n+1. 

Whence, 

•    g  _  q0  4-(«i  4-  P«o)a  +(pqn-i  +  gan-2)xn  +  gan-izw+1 .  ^n 

1  +  px  4-  ##2 

which  is  a  formula  for  the  sum  of  the  first  n  terms  of  a  recurring  series  of 
the  second  order. 

If  x  is  so  taken  that  the  given  series  is  convergent,  xn  and  xn+l  approach 
the  limit  0,  when  n  is  indefinitely  increased,  and  the  fraction  (1)  ap- 
proaches the  limit 

1  +  px  4*  qx2 

If  this  fraction  be  expanded  into  an  infinite  series  by  division,  we 
obtain  the  given  series  ;  but  it  is  only  when  the  series  is  convergent  that 
it  expresses  the  value  of  the  fraction. 


INFINITE   SERIES  185 

Then,  the  sum  of  the  given  series  (§  250)  is  given  by  the  formula 

# -ftp  +Oi  4- p«n)a;i  /2) 

1  +  px  -f  qxl 

If  q  =  0,  the  series  is  of  the  first  order,  and  «i  +p«o  =  0  ;  then 

1  +  px 

which  is  a  formula  for  the  sum  of  a  recurring  series  of  the  first  order. 
(Compare  §  242.) 

In  like  manner,  we  shall  find  the  formula 

8  _  tto  4-  (fli  4-  pao)x  4-  (ff2  +  ff«i  4-  gao)x2  ^4n 

1  +px  4-  gx2  4-  rx3 

for  the  sum  of  a  recurring  series  of  the  third  order. 

255.  A  recurring  series  is  formed  by  the  expansion  in  an 
infinite  series  of  a  fraction,  called  the  generating  fraction. 

The  operation  of  summation  reproduces  the  fraction,  the 
process  being  the  reverse  of  that  of  §  26S. 

256.  Ex.   Find  the  sum  of  the  series 

2  +  x  +  5x2  +  7a?  +  17x4+  .... 

To  determine  the  scale  of  relation,  we  first  assume  the  series  to  be  of 
the  second  order  (§  253). 

Substituting  a0  -  2,  a\  —  1,  a2  =  5,  as  —  7,  in  (1),  §  253, 


J  5  +    p+2q=0, 

1  7  +  rop  4-     q  =  Q. 


Solving  these  equations,  p—  —  1,  tf  —  -2. 

To  ascertain  if  1  —  x  —  2  x2  is  the  correct  scale  of  relation,  consider  the 
fifth  term. 

Since  17  x4  +  (-  x)  (7  x3)  +  (-  2  x2)  (5  x2)  is  equal  to  0,  it  follows  that 
1  —  x  —  2  x2  is  the  correct  scale. 

Substituting  the  values  of  a0,  «i,  i>,  and  q  in  (2), 

jy_24(l-2)g_       2-x 

l-x-2x2       l-x-2x2' 

The  result  may  be  verified  by  expansion. 


186  ALGEBRA 

EXERCISE   63 

Find  the  sum  of  the  following: 

l  4-r-#  +  7#2  — 5ar3+19^H . 

2.  l-13x-23x2-85xi-239x4+  .... 

3.  l+5x  +  21x2  +  85x3  +  Mlxi+  .... 

4.  5-13a  +  35a2-97ar3-h275a4+  .... 

5.  3  +  10#  +  36^  +  136arJ  +  528a4+ .... 

6.  3  +  x  +  33x2  +  109x*  +  657xi+  .... 

7.  14. 2  #-  3 a;2  +  6^r3  -7  a;4  +  10 a;5  -11  a6  +  .... 

8.  l-2x-a2-7ar3-18#4-59;r5-181a6  +  .... 

THE   DIFFERENTIAL   METHOD 

257.  If  the  first  term  of  a  series  be  subtracted  from  the 
second,  the  second  from  the  third,  and  so  on,  the  series  formed 
is  called  the  first  order  of  differences  of  the  given  series. 

The  first  order  of  differences  of  this  new  series  is  called 
the  second  order  of  differences  of  the  given  series ;  and  so  on. 

Thus,  in  the  series 

1,     8,     27,     64,     125,     216,      .., 

the  successive  orders  of  differences  are  as  follows  : 

1st  order,  7,     19,     37,    61,  .  91,     .... 

2d   order,  12,     18,     24,     30,     .... 

3d  order,  6,       6,       6,     .... 

4th  order,  0,        0,     .... 

The  Differential  Method  is  a  method  for  finding  any  term,  or 
the  sum  of  any  number  of  terms  of  a  series,  by  means  of  its 
successive  orders  of  differences. 

258.  To  find  any  term  of  the  series 

aU        a2)        a3)        a4)        '")        a»l        an+l>        '"• 

The  successive  orders  of  differences  are  as  follows: 
1st  order,     a2  —  «i,  03  —  «2,  «4  —  «8i  •••»  «n+i  —  «n,  •••• 
2d  order,     a3  —  2a»  +  <*h  a*  —  2  a*  +  a^  '"- 
3d  order,     a4  —  3  a3  +  3  a2  —  «i>  •••  J  etc. 


INFINITE   SERIES  187 

Denoting  the  first  terms  of  the  1st,  2d,  3d,  •••,  orders  of  dif- 
ferences by  dly  d2,  d3,  •••,  respectively,  we  have 

di  =  a2  —  a\ ;  whence,  a2  =  «i  +  d\. 

d2  =  as  —  2  a2  +  «i ;  whence, 

as  =  —  «i  +  2  ^2  +  d2  =  —  a\  +  2  «x  +  2  c?i  +  d2  =  cti  +  2  dfi  +  (^. 

e?3  =  «4  —  3  a3  +  3  a2  —  «i ;  whence, 

«4  =  «i  —  3  a2  +  3  a3  +  (?3  =  «i  -f-  3  di  +  3  d2  +  d3 ;  etc. 

It  will  be  observed,  in  the  values  of  a2,  a3,  and  a4,  that  the 
coefficients  of  the  terms  are  the  same  as  the  coefficients  of  the 
terms  in  the  expansion  by  the  Binomial  Theorem  of  a -J- a;  to 
the  first,  second,  and  third  powers,  respectively. 

We  will  now  prove  by  Mathematical  Induction  that  this  law 
holds  for  any  term  of  the  given  series. 

Assume  the  law  to  hold  for  the  ?ith  term,  an ;  then  the  coef- 
ficients of  the  terms  will  be  the  same  as  the  coefficients  of  the 
terms  in  the  expansion  by  the  Binomial  Theorem  of  a  +  x  to 
the  (n  —  l)th  power ;  that  is, 

an=a1+(n-l)d1  +  ^^^^ld2 

+  (n-l)(»  -2)^-3)^^  (1) 

lit 
If  the  law  holds  for  the  nth  term  of  any  series,  it  must  also 
hold  for  the  nth  term  of  the  first  order  of  differences ;  or, 

an+1-an  =  dl  +  (n-l)d2+(n-1^n-2^d3+  ....        (2) 

Adding  (1)  and  (2),  we  have 

an+1  =  a1  +  [(n-l)+iyi1  +  1^±[(n-2)  +  2]d2 

+  »-1^2)[(n-3)  +  3]d<+... 

=  fll  +  ndl  +  !l^^  (3) 

If.  l£ 

This  result  is  in  accordance  with  the  above  law. 


188  ALGEBRA 

Hence,  if  the  law  holds  for  the  nth.  term  of  the  given  series, 
it  holds  for  the  (n  +  l)th  term ;  but  we  know  that  it  holds  for 
the  fourth  term,  and  hence  it  holds  for  the  fifth  term;  and 
so  on. 

Therefore,  (1)  holds  for  any  term  of  the  given  series. 

If  the  differences  finally  become  zero,  the  value  of  an  can  be  obtained 
exactly. 

259.  To  find  the  sum  of  the  first  n  terms  of  the  series 

Ctiy   0,9,   &3,   &4,   &5,    ••*.  {±J 

Let  S  denote  the  sum  of  the  first  n  terms. 
Then  S  is  the  (n  +  l)th  term  of  the  series 

0,  ft,  ai  +  «2,  cii  +  a2  +  as,  ••••  (2) 

The  first  order  of  differences  of  (2)  is  the  same  as  (1)  ;  whence,  the 
2d  order  of  differences  of  (2)  is  the  same  as  the  1st  order  of  differences 
of  (I),  the  3d  order  of  (2)  is  the  same  as  the  2d  order  of  (1),  and  so  on. 

Then,  if  di,  d2,  •••,  represent  the  first  terms  of  the  1st,  2d,  •  •-,  orders 
of  differences  of  (1),  cti,  di,  d2,  ...,  will  be  the  first  terms  of  the  1st,  2d, 
3d,  •••,  orders  of  differences  of  (2). 

Putting  a\  —  0,  d\  =  «i,  d2  —  c?i,  etc.,  in  (3),  §  258, 

<y=nfl1  +  <wt"1^i  +  <w"1^w"2>da  +  -.  (3) 

\1 .  12. 

260.  Ex.  Find  the  twelfth  term,  and  the  sum  of  the  first 
twelve  terms,  of  the  series  1,  8,  27,  64,  125,  •••. 

Here,  n  =  12,  a\  xs  1. 

Also,  di  =  7,  d2  =  12,  <h  =  0,  and  d4  =  0  (§  257). 

Substituting  in  (1),  §  258,  the  twelfth  term 

a- 1  +  11 . 7  +  llll^  .  12  +  11,10,9  .  6  =  1728. 
1-2  1.2-3 

Substituting  in  (3),  §  250,  the  sum  of  the  first  twelve  terms 

=  12  +  l^H.7  +  ]2-11-10.12  +  12-11-:l0-i>-0  =  C084. 
1.2  1.2-3  1.2-3.4 


INFINITE    SERIES  189 

261.    Piles  of  Shot. 

Ex.  If  shot  be  piled  in  the  shape  of  a  pyramid  with  a  tri- 
angular base,  each  side  of  which  exhibits  9  shot,  find  the  num- 
ber in  the  pile. 

The  number  of  shot  in  the  first  five  courses  are  1,  3,  6,  10,  and  15, 
respectively  ;  we  have  then  to  find  the  sum  of  the  first  nine  terms  of  the 
series  1,  3,  6,  10,  15,  •  ••  ;  the  successive  orders  of  differences  are  as 
follows : 

1st  order,  .  2,     3,     4,     5,     .... 

2d  order,  .  1,     1,     1,     .... 

3d  order,  .  0,     0,     .... 

Putting  n  =  9,  d!  =  1,  di  =  2,  d2  =  1  in  (3),  §  259, 

S  =  9  +  —  •  2  +  9'8'7  •  1  =  165  . 
1-2  1-2-3 

.  EXERCISE   64 

i.  Find  the  first  term  of  the  sixth  order  of  differences  of 
the  series  3,  5,  11,  27,  67,  159,  375,  •  ••. 

2.  Find  the  15th  term,  and  the  sum  of  the  first  15  terms,  of 
the  series  1,  9,  21,  37,  57,  •••. 

3.  Find  the  14th  term,  and  the  sum  of  the  first  14  terms,  of 
the  series  5,  14,  15,  8,  —7,  •••. 

4.  Find  the  sum  of  the  first  n  multiples  of  3. 

5.  If  shot  be  piled  in  the  shape  of  a  pyramid  with  a  square 
base,  each  side  of  which  exhibits  25  shot,  find  the  number  in 

the  pile.  < 

6.  Find  the  13th  term,  and  the  sum  of  the  first  13  terms,  of 
the  series  1,  3,  9,  25,  57,  111,  .... 

7.  Find  the  10th  term,  and  the  sum  of  the  first  10  terms,  of 
the  series  4,  -  2,  10,  4,  -  56,  -  206,  .... 

8.  Find  the  sum  of  the  squares  of  the  first  n  multiples  of  2. 

9.  Find  the  71th  term,  and  the  sum  of  the  first  n  terms,  of 
the  series  1,  —  3,  — 13,  —  17,  —3,  41,  .... 


190  ALGEBRA 

10.  Find  the  number  of  shot  in  a  pile  of  9  courses,  with  a 
rectangular  base,  if  the  number  of  shot  in  the  longest  side  of 
the  base  is  24. 

ii.  Find  the  number  of  shot  in  a  truncated  pile  of  8 
courses,  with  a  rectangular  base,  if  the  number  of  shot  in 
the  length  and  breadth  of  the  base  are  20  and  14,  respec- 
tively. 

12.  Find  the  12th  term,  and  the  sum  of  the  first  12  terms,  of 
the  series  1,  13,  49,  139,  333,  701,  1333,  .... 

13.  Find  the  9th  term,  and  £he  sum  of  the  first  9  terms,  of 
the  series  20,  4,  -36,  -132,  -356,  -820,  -1676,  .... 

14.  Find  the  sum  of  the  fourth  powers  of  the  first  n  natural 
numbers. 

15.  Find  the  number  of  shot  in  a  pile  with  a  rectangular 
base,  if  the  number  of  shot  in  the  length  and  breadth  of  the 
base  are  m  and  n,  respectively. 

16.  How  many  shot  are  contained  in  a  truncated  pile  of  n 
courses,  whose  bases  are  triangles,  if  the  number  of  shot  in 
each  side  of  the  upper  base  is  m  ? 


INTERPOLATION 

262.  Interpolation  is  the  process  of  introducing  between  the 
terms  of  a  series  other  terms  conforming  to  the  law  of  the 
series. 

Its  usual  application  is  in  finding  intermediate  numbers  be- 
tween those  given  in  Mathematical  Tables. 

The  operation  is  effected  by  giving  fractional  values  to  n  in 
(1),  §  258. 

The  method  of  Interpolation  rests  on  the  assumption  that  a 
formula  which  has  been  proved  for  an  integral  value  of  n, 
holds  also  when  n  is  fractional. 


INFINITE   SERIES  191 

263.   Ex.     Given  V5  =  2.2361,  V0  =  2.4495,   V7  =  2.6458, 
V8  =  2.8284,  •  •  • ;  find  Vo\3. 
In  this  case  the  successive  orders  of  differences  are  : 
.2134,     .1963,     .1826,      ... 
-.0171, -.0137,     .... 
.0034,     .... 

Whence,  dx  =  .2134,  d2=-.0171,  d3  =  .0034,  .... 
Now,  the  required  term  is  distant  1.3  intervals  from  V5. 
Substituting  n  =  2.3  in  (1),  §  258,  we  have,  approximately, 

V6\3  =  2.2361  + 1.3  x  .2134  +  1'3  x  f  ( -  .0171) 

1.3  x  .3x-.7      >0034 
1x2x3 

=  2.2361  +  .2774  -  .0033  -  .0002  =  2.5100. 


EXERCISE   65 

i.  Given  log  26  =  1.4150,  log  27  =  1.4314,  log  28  =  1.4472, 
log  29  =  1.4624,  • . . ;  find  log  26.7. 

2.  Given  r#$[  =  4.49794,  ^92  =  4.51436,  ^93  =  4.53066, 
^94  =  4.54684,  •  -.;  find  y/WS. 

3.  The  reciprocal  of  35  is  .02857;  of  36,  .02778;  of 
37,  .02703;  of  38,  .02632;  etc.  Find  the  reciprocal  of 
36.28, 

4.  Given  log  124  =  2.09342,  log  125  =  2.09691,  log  126 
=  2.10037,  log  127  =  2.10380,  •  • . ;  find  log  125.36. 

5.  Given  213  =  9261,  228  =  10648,  233  =  12167,  243  =  13824, 
and  253  =  15625 ;  find  the  cube  of  21f 

6.  Given  log  61  =  1.78533,  log  62  =  1.79239,  log  63  =  1.79934, 
log  64  =  1.80618,  ... ;  find  log  63,527, 


192  ALGEBRA 

XI.     UNDETERMINED   COEFFICIENTS 

THE   THEOREM   OF   UNDETERMINED    COEFFICIENTS 

264.  An  important  method  for  expanding  expressions  into 
series  is  based  on  tLe  following  theorem : 

If  the  series  A  +  Bx  +  Cx2  +  Dx*  +  •••  is  always  equal 
to  the  series  A'  +  B'x  +  Cx2  +  D'x3  +  •••,  when  x  has  any 
value  which  makes  both  series  convergent,  the  coef- 
ficients of  like  powers  of  x  in  the  series  will  be  equal ; 
that  is,  A  =  A',  B  =  B<,  C  =  C. 

265.  Before  giving  the  proof  of  the  Theorem  of  Undeter- 
mined Coefficients,  we  will  prove  two  theorems  in  regard  to 
infinite  series. 

First,  if  the  infinite  series 

a  +  bx  +  cx2  +  dx2  -f  ••• 

is  convergent  for  some  finite  value  of  x,  it  is  finite  for  this 
value  of  x  (§  247),  and  therefore  finite  when  x  =  0. 

Hence,  the  series  is  convergent  when  x  =  0. 

Second,  if  the  infinite  series 

ax  +  bx2  +  ex5  +  ••• 

is  convergent  for  some  finite  value  of  x,  it  equals  0  when  x  =  0. 

For,  ax  +  bx2 -f  ex3  +  •••  is  finite  for  this  value  of  x,  and 
hence  a  +  bx  +  ex2  +  •  •  •  is  finite  for  this  value  of  x. 

Then,  a  +  bx  +  cx2-\-  ..«  is  finite  when  #  =  0;  and  therefore 
»(a  +  bx  +  cx2+  •••),  or  ax  +  bx2  +  cx3+  •••,  equals  0  when 
x  =  0. 

266.  Proof  of  the  Theorem  of  Undetermined  Coefficients. 

The  equation 

A  +  Bx+Cx2+Dx*+  ...  =A'  +  B'x+C'x°-  +  D'x»+  ...     (1) 

is  satisfied  when  x  has  any  value  which  makes  both  members 
convergent;  and  since  both  members  are  convergent  when 
x  =  0  (§  265),  the  equation  is  satisfied  when  x  =  0. 


UNDETERMINED   COEFFICIENTS  193 

Putting  x  =  0,  we  have,  by  §  265, 

Bx+Cx*  +  Dx*  +  .-•  =0,  and  B'x  +  Ox2  +  DW  +  •••  =0. 

Whence,  A  =  A'. 

Subtracting  A  from  the  first  member  of  (1),  and  its  equal  A' 
from  the  second  member,  we  have 

Bx+Cx*  +  Dx*  +  ...=B'x'+C'x2  +  D'x'  +  .... 

Dividing  each  term  by  x, 

B+Cx  +  Dx2+  ...  =B'  +  Cx  +  D'x*+  ....  (2) 

The  members  of  this  equation  are  finite  for  the  same  values 
of  x  as  the  given  series  (§  265). 

Then,  they  are  convergent,  and  therefore  equal,  for  the  same 
values  of  x  as  the  given  series. 

Then  the  equation  (2)  is  satisfied  when  x  =  0. 

Putting  x  =  0,  we  have  B  =  B'. 

Proceeding  in  this  way,  we  may  prove  C=  (7,  etc. 

267.  The  theorem  of  §  264  holds  when  either  or  both  of  the 
given  series  are  finite. 

EXPANSION   OP   FRACTIONS 

2 3  x2  —  x3 

268.  i.   Expand in  ascending  powers  of  x. 

1  —  2  x  -f  3  x1 

Assume  2  ~  8  ^  ~  ^  =  A  +  Bx  +  Cx2  +  B&  +  Ex*  +  — ,  (1) 

1  —  2  x  +  3  x2 

where  ^4,  J5,  0,  Z>,  E,  •••,  are  numbers  independent  of  x. 

Clearing  of  fractions,  and  collecting  the  terms  in  the  second  member 
involving  like  powers  of  x,  we  have 

*4+-.        (2) 


2-Sx2-x3  =  A+     B\x+     C 
-2A\     -2B 


*:2+  D 
-2C 
+  3B 


x3+     E 
-2D 

4-3(7 


A  vertical  line,  called  a  bar,  is  often  used  in  place  of  parentheses. 

Thus,        "  +     B  I  x  is  equivalent  to  (B  —  2  A)x. 

-2,4 


194  ALGEBRA 

The  second  member  of  (1)  must  express  the  value  of  the  fraction  for 
every  value  of  x  which  makes  the  series  convergent  (§  249)  ;  and  there- 
fore equation  (2)  is  satisfied  when  x  has  any  value  which  makes  the 
second  member  convergent. 

Then,  by  §  267,  the  coefficients  of  like  powers  of  x  in  (2)  must  be 
equal ;  that  is, 

A=     2. 
B-2A  =      0;0r,  £  =  2.4  =  4. 

C-2B  +  3A  =  -S;  or,  C  =  2  B-  3  A  -  3  =  .-  1. 
Z>_2  C+3.B=- 1;  or,  Z)  =  2  C- 3  2?  -  1  =  -  15. 
E-2D+3C=     0;or,  E=2D-3C        =-27;  etc. 

Substituting  these  values  in  (1),  we  have 

2  _  3  x2  -  x 


1  —  2  x  +  3  x2 


=  2+4x-x2-\5xS-27x* . 


The  result  may  be  verified  by  division. 

The  series  expresses  the  value  of  the  fraction  only  for  such  values  of  x 
as  make  it  convergent  (§  249). 

If  the  numerator  and  denominator  contain  only  even  powers 
of  x,  the  operation  may  be  abridged  by  assuming  a  series  con- 
taining only  the  even  powers  of  x. 

2  -I-  4  a?2 x4 

Thus,  if  the  fraction  were  — — — ■ ,  we  should  assume 

it  equal  to  A  +  B  x2  +  C  xA  +  D  xG  +  E  ars  -f  — . 

In  like  manner,  if  the  numerator  contains  only  odd  powers 
of  05,  and  the  denominator  only  even  powers,  we  should  assume 
a  series  containing  only  the  odd  powers  of  x. 

If  every  term  of  the  numerator  contains  x,  we  may  assume  a 
series  commencing  with  the  lowest  power  of  x  in  the  numerator. 

If  every  term  of  the  denominator  contains  x,  we  determine 
by  actual  division  what  power  of  x  will  occur  in  the  first  term 
of  the  expansion,  and  then  assume  the  fraction  equal  to  a  series 
commencing  with  this  power  of  x,  the  exponents  of  x  in  the 
succeeding  terms  increasing  by  unity  as  before. 

2.   Expand  — in  ascending  powers  of  x. 

3  x  -—  xr 


undeterminp:d  coefficients  195 


Dividing  1  by  3  ic2,  the  quotient  is  — - ;  we  then  assume, 

o 

=  Ax-'2  +  Bx~*  +  C  +  Dx  +  Ex2  +  — .  (3) 


3  x2  -  x3 
Clearing  of  fractions. 


l=3^  +  3#|x  +  3C|x2  +  3.D|z3  +  3.E' 
_    A      -     i?       -     C       -     i> 


z4  +  « 


Equating  coefficients  of  like  powers  of  x, 

SA  =  \,3B-A  =  013C-B  =  0,3D-C  =  0,3E-D  =  0',  etc. 

Whence,     A  =  \,  B=l,   0  =  1,  D  =  ±,  **^.~. 

Substituting  in  (8),  _£_  -^f|  +  i  +  Jf^*- 

In  Ex.  1,  E  =  2  D  —  3  C ;  that  is,  the  coefficient  of  se4  equals  twice  the 
coefficient  of  the  preceding  term,  minus  three  times  the  coefficient  of  the 
next  but  one  preceding. 

It  is  evident  that  this  law  holds  for  the  succeeding  terms  ;  thus,  the 
coefficient  of  x5  is  2  x  (-  27)  -  3  x  (-  15),  or  -  9. 

After  the  law  of  coefficients  has  been  found  in  any  expansion,  the  terms 
may  be  found  more  easily  than  by  long  division  ;  and  for  this  reason  the 
method  of  §  268  is  to  be  preferred  when  a  large  number  of  terms  is 
required. 

The  law  for  Ex.  2  is  that  each  coefficient  is  one-third  the  preceding. 

EXERCISE   66 

Expand  each  of  the  following  to  five  terms  in  ascending 
powers  of  x : 

4  +  2a  3  +  x  +  x2  1  +  2  a;  +  4  .r2 

'    l-Vx  +  3-x*' 

5  4-  .6  x2 
2-3x2  +  x3' 

1  _  6  tf  +  4  g8 
z+Jf^2& 


1-23 

l-8aj 

l  +  5x 

2  +  x2 

1-2  x2 

5  x 

5+2^-7^ 

4  +  X-3X2 

1-2Z2 

x  _  5  x*  4. 8  x4 

2  _  3  .T  +  z2 

4  x2  -  3  .t3 

10. 


7.   ~ — : » 11. 


4  8  12 

1-r-Gar9'  '    6-4z-5ar3'  '    2rJ  +  4ar5  +  a6 


196 


ALGEBRA 

EXPANSION   OF   SURDS 


269.   Ex.   Expand  Vl  —  x  in  ascending  powers  of  x. 
Assume  Vl  -  x  =  A  +  Bx  +  Cx'2  -f  Dx*  +  E&  +  •  ••. 

Squaring  both  members,  we  have,  by  §  167, 


(1) 


l-x  =  A* 


+  2AB 


x+      B2\x2 
+  2AC\      +  2AD 


x*+  C2 
+  2AE 
+  2BD 


\      +  2BC 

Equating  coefficients  of  like  powers  of  x, 

A2  =      1;  or,  A=l. 

2AB  =  -1;   or,  B=-  _I=-i. 
2A  2 


x*  +  - 


0;  or,  C=- 


&  +  2AC 
2AD+2BC 
C2  +  2AE  +  2BD  =      0  ;  or,  E 


2A 


0;  or,  B=-^  = 
A 


1 

"8* 

t 

16* 


C2  +  2BD_ 


2A 


128' 


etc. 


Substituting  these  values  in  (1),  we  have 

x _  x? _ x*_  _  Sac4 

VI  -  x  =  1  -  2       g       16      128  " 

The  result  may  be  verified  by  Evolution. 


The  series  expresses  the  value  of  Vl  —  x  only  for  such  values  of  x  as 
make  it  convergent. 

EXERCISE   67 

Expand  each  of   the  following  to  five   terms  in  ascending 
powers  of  x: 


i.    Vl  +  2  x. 
2.    Vl  -  3  X. 


3-    Vl-4a  +  a;2.        5.    Vl-f6z. 

4.    Vl  -f  x  —  x2.  6.   Vl  —  x  —  2  a:-. 


PARTIAL,   FRACTIONS 

270.   If  the  denominator  of  a  fraction  can  be  resolved  into 
factors,  each  of  the  first  degree  in  x}  and  the  numerator  is  of  a 


UNDETERMINED   COEFFICIENTS  197 

lower  degree  than  the  denominator,  the  Theorem  of  Undeter- 
mined Coefficients  enables  us  to  express  the  given  fraction  as 
the  sum  of  two  or  more  partial  fractions,  whose  denominators 
are  factors  of  the  given  denominator,  and  whose  numerators 
are  independent  of  x. 

271.    Case  I.     No  factors  of  the  denominator  equal. 

19  x  4- 1 

i.    Separate  — — — into  partial  fractions. 

1  (3x-l)(5x  +  2)  ' 

Assume  19*+1 =      A      +      B     ,  (1) 

where  A  and  B  are  numbers  independent  of  as. 

Clearing  of  fractions,     19  x  +  1  =  A(JS  x  +  2)  -f  B(?>  x  —  1). 

Or,  19x+l  =  (5A+SB)x  +  2A-B.  (2) 

The  second  member  of  (1)  must  express  the  value  of  the  given  fraction 
for  every  value  of  x. 

Hence,  equation  (2)  is  satisfied  by  every  value  of  x ;  and  by  §  267,  the 
coefficients  of  like  powers  of  x  in  the  two  members  are  equal. 

That  is,  5i  +  35  =  19, 

and  2A-B  =  1. 

Solving  these  equations,  we  obtain  A  =  2  and  B  =  3. 

Substituting  in  (1), 19  x  +  1 = ? l ? 

(8  x  -  1)  (6  x  +  St;      3  x  -  1      5  x  +  2 

The  result  may  be  verified  by  finding  the  sum  of  the  partial 
fractions. 

2.    Separate -^ into  partial  fractions. 

2x  —  x~  —  x3 

The  factors  of  2  a  -  x2  -  %*  are  sc,  1  -  x,  and  2  +  x  (§  103,  III,  VIII). 

Assume  then,       x+A —  _ A  +  _JL_ _( *?_i. 

2  x  —  x2  —  x3       x      1  —  x     2  +  x 

Clearing  of  fractions,  we  have 

x  +  4  =  ^4(1  -x)(2  +  x)  +  Bx(2  +  x)  +  Cx(l  -  x). 

This  equation,  being  satisfied  by  every  value  of  x,  is  satisfied  when  x  =  0* 


198  ALGEBRA 

Putting  x  =  0,  we  have  4  =  2  A,  or  A  —  2. 

Again,  the  equation  is  satisfied  when  x  =  1. 

5 
Putting  x  =  1,  we  have  5  =  3  J5,  or  i?  =  -- 

3 

The  equation  is  also  satisfied  when  x.=  —  2. 

Putting  x  =  —  2,  we  have  2^—6  C,  or  C  =  —  J. 

The„,        *  +  4 =  ?  +  _L  +  ^i_  =  ?  + 6 1 

'  2x  —  x2  —  xs     x      1  —  x     2  +  x     x      S(l  —  x)      3(2  +  x) 

To  find  the  value  of  A,  in  Ex.  2,  we  give  to  x  stick  a  value  as  will 
make  the  coefficients  of  B  and  C  equal  to  zero  ;  and  we  proceed  In  a 
similar  manner. to  find  the  values  of  B  and  C. 

This  method  of  finding  A,  B,  and  C  is  usually  shorter  than  that  used 
in  Ex.  1. 


Case  II.     All  the  factors  of  the  denominator  equal, 

x*  -  jla?  +  26  . 


Let   it  be   required   to   separate — — -~ —  into  partial 


fractions. 

Substituting  y  -f  3  for  a;,  the  fraction  becomes 

(y  +  3)2  -  ll(y  +3)+26  =  y2-5y  +  2  =  l       5   |   2 

2/3  2/3  y    y'2    y* 

Replacing  y  by  x  —  3,  the  result  takes  the  form 
1  5  2 


as  _  3      (a  _  3)2      (x  -  3)8 

This  shows  that  the  given  fraction  can  be  expressed  as  the  sum  of 
three  partial  fractions,  whose  numerators  are  independent  of  .r,  and 
whose  denominators  are  the  powers  of  x  —  3  beginning  with  the  first  and 
ending  with  the  third. 

Similar  considerations  hold  with  respect  to  any  exam  pic1 
under  Case  II ;  the  number  of  partial  fractions  in  any  case 
being  the  same  as  the  number  of  equal  factors  in  the  denomi- 
nator of  the  given  fraction. 

6  x  4-  5 
Ex.     Separate — —  into  partial  fractions. 

(3x  +  5)2  l 

In  accordance  with  the  above  principle,  we  assume  the  given  fraction 


UNDETERMINED   COEFFICIENTS  199 

equal  to  the  sum  of  two  partial  fractions,  whose  denominators  are  the 
powers  of  8  x  +  5  beginning  with  the  first  and  ending  with  the  second. 

That  is,  6*  +  5,=      A      +-      B 


Ex.     Separate  — - —     ^~2  ■  into  partial  fractions. 


(3x  +  5)2      3x  +  5      (3x  +  5)2 
Clearing  of  fractions,  6  x  +  5  =  4.(3  a;  +  5)  +  B. 
=  SAx  +  5A  +  B. 
Equating  coefficients  of  like  powers  of  x, 
3  A  =  6, 
and  5  A  +  5  =  5. 

Solving  these  equations,      J.  =  2  and  i?  =  —  5. 

Whence  _6x±6_  =  _2 5 

(3  x  +  5)2     3  x  +  5      (3  x  +  5)2 

Case  III.     Some  of  the  factors  of  the  denominator  equal. 

a?  -  4  a?  -f  3  .. 
a?(a?  +  l)2 

The  method  in  Case  III  is  a  combination  of  the  methods  of  Cases  I  and 
II ;  we  assume, 

x2  -  4  x  +  3  __  A         B  C 

x(x  +  l)2        x      x  + 1      (x  +  l)2' 

Clearing  of  fractions, 

x2  — 4x  +  3  =  4(x  +  l)2+ J3x(x  +  jj+  0a. 

=  (A  +  B)x2  +  (2A  +  B+C)x  +  A. 
Equating  coefficients  of  like  powers  of  x, 
A  +  B  =  l, 
2A+  J5+  C  =  -4, 
and  A  =  3. 

Solving  these  equations,  A  =  3,  J5  =  —  2,  and  (7  =  —  8. 

Whence,  *-**  +  «  =  ?  _  _^_  _        8       . 

x(x  +  l)2       X      x+1      (x  +  1)2 

The  following  general  rule  for  Case  III  will  be  found  convenient : 

X 

A  fraction  of  the  form should  be  assumed 


equal  to  {x  +  a)(.x+b)-(.x  +  my 


*  .+     *    +...  +  ^  +  _JL_ +  ...+  _^_+.. 


x  +  a     x  +  b  x  +  m      (x  +  m)2  (x  +  m)r 


200  ALGEBRA 

single  factors  like  x  +  a  and  x  +  b  having  single  partial  fractions  cor- 
responding, arranged  as  in  Case  I ;  and  repeated  factors  like  (x  +  m) 
having  r  partial  fractions  corresponding,  arranged  as  in  Case  II. 

272.  If  the  degree  of  the  numerator  is  equal  to,  or  greater 
than,  that  of  the  denominator,  the  preceding  methods  are 
inapplicable. 

In  such  a  case,  we  divide  the  numerator  by  the  denominator 
until  a  remainder  is  obtained  which  is  of  a  lower  degree  than 
the  denominator. 

Xs  _  gx  2  _  j 
Ex.    Separate into  an  integral  expression  and 

partial  fractions. 

Dividing  x8  —  3  x2  —  1  by  x2  —  x,  the  quotient  is  x  —  2,  and  the  re- 
mainder —  2  x  —  1 ;  we  then  have 

a*-3a*-l=a;     2  ,    -2S-1,  (1) 

X2  —  X  X2—  X 
O  y i 

We  can  now  separate into  partial  fractions  by  the  method 

x2  —  x 

1         3 
of  Case  I ;  the  result  is 


x     x—  1 

Substituting  in  (1), ^ i  =  x_2  +  ---   e 


x     x  —  1 

Another  way  to  solve  the  above  example  is  to  combine  the  methods  of 
§§  268  and  271,  and  assume  the  given  fraction  equal  to 

Ax  +  B  +  C  +  _D^. 
x      x—  1 

273.  If  the  denominator  of  a  fraction  can  be  resolved  into 
factors  partly  of  the  first  and  partly  of  the  second,  or  all  of  the 
second  degree,  in  x,  and  the  numerator  is  of  a  lower  degree 
than  the  denominator,  the  Theorem  of  Undetermined  Coeffi- 
cients enables  us  to  express  the  given  fraction -as  the  sum  of 
two  or  more  partial  fractions,  whose  denominators  are  factors 
of  the  given  denominator,  and  whose  numerators  are  inde- 
pendent of  x  in  the  case  of  fractions  corresponding  to  factors 


UNDETERMINED   COEFFICIENTS  201 

of  the  first  degree,  and  of  the  form  Ax  +  B  in   the   case    of 
fractions  corresponding  to  factors  of  the  second  degree. 

The  only  exceptions  occur  when  the  factors  of  the  denominator  are  of 
the  second  degree  and  all  equal. 

Ex.    Separate into  partial  fractions. 

x3  +  l 

The  factors  of  the  denominator  are  x  +  1  and  x2—  x  +  1. 

Assume  then  -1—  =  -A_  +    fx +V  .  (1) 

x*  +  l      x+l     x2-x+l 

Clearing  of  fractions,       1  =  A(x2  —  x  +  1)  +  (Bx  +  C)  (x  +  1). 

Or,  %  ^  (4  +  B)x2  +  (-  A  +  B  +  C)x  +  A  +  C. 

Equating  coefficients  of  like  powers  of  x, 

A  +  B  =  0, 

-A  +  B+C  =  0, 

and  .4  +  C  =  1. 

Solving  these  equations,  A  =  J,  I?  =  —  J,  and  (7  =  f. 
1  1  x-2 


Substituting  in  (1), 


x3  +  l     8(x  +  l)      3(£2-x+l) 


EXERCISE   68 

Separate  into  partial  fractions : 
%  -1  ■       6      a8  +  4a2  +  2a;  +  3 


a?  _  9  3  +  20  (a;2  + 1)  (a;2  +  x  +  1) 

15x-2T  „         3a;-7 

7- 


10x2  +  ^-21  ^-2^-8 

2a?  +  3  8        6x?-12 


a-*-a&-12  a^-5a;2-f-4 

12  a;  + 18  «2~15a?  +  3 

^  +  3a2-18x*  ar9-3x-28' 

43a-31  5a2  +  16a;-2 


30  a2 -12  a; -306  ^  +  4^-3^-18 


202 


ALGEBRA 


12. 


13. 


14. 


18. 


Sxi-^16xz-10x2-^2Sx-\-ll 
"*  2x*  +  x-3 

59x-5S 


12  a;2 -25  a +  12* 

2^-11^4- 19 
x*-§x*  +  12x-% 

2xs-3x-8 
(x>  +  x-2)2'     . 

X3—  X 


15. 


16. 


17. 


2x*  +  x2  +  5x 
(xi  +  2x  +  l)(x2-x  +  l) 

x*  +  2a?-9x2  +  7x 
x4  —  4:Xs  +  6x2  —  4:X  +  l 
12ft4  +  19ar*-7a: 


4a;4 +  1 

2ar>-2 


5L-? 19.     **-*.     20 

x(x2+xjr\)+2(x2+x+l)  #4+a;2+-l 


2x-Z 
4tx*—x 


REVERSION   OP   SERIES 

274.   To  revert  a   given   series  y  =  a  +  bxm  +-  cxn  +-  •••  is   to 
express  #  as  a  series  proceeding  in  ascending  powers  of  y. 

Ex.   Revert  the  series  y  =  2x  —  3x2-\-4:Xs  —  5x4  +  •••. 

Assume  x  =  Ay  +  By2  +  Cyz  +  Zty4  +  •••. 

Substituting  in  this  the  given  value  of  y, 

x  =  A(2  x  -  3  x2  +  4x3  -  5  z4  +  •••) 

+  5(4  x2  +  9  £4  -  12  x3  +  16  x*  +  — ) 
-f  (7(8  x3  -  36  x4  +  .-•)  +  2>(16  x4  +  ■ 


(1) 


)+" 


That  is,  x  =  2  ^4x  -  3  .4 1  x2  +    4  .4 

+  45!      -  12  B 

■      +    8(7 


x3  —  5  ^4 
+  25  5 
-36(7 
+  16  2) 


sc4  + 


Equating  coefficients  of  like  powers  of  x, 

2.4  =  1; 

4  4-125  +  8  (7=0; 
-  5  .4  +  25  B  -  36  (7  +  16  D  =  0  ;  etc. 
Solving,  A  s  J,  5  =  f ,  (7  =  Ai  *>  =  f¥s,  etc. 

Substituting  in  (1) ,        x  =  J  y  +  f  y2  +  ^  y3  +  T32\  y4  +  •  •  -. 

If  the  even  powers  of  x  are  wanting  in  the  given  series,  the 
operation  may  be  abridged  by  assuming  x  equal  to  a  series 
containing  only  the  odd  powers  of  y. 


pp:rmutations  and  combinations  203 

EXERCISE   69 

Revert  each  of  the  following  to  four  terms : 
i.   2/  =  ^  +  3^  +  5x3  +  7^+  ....     3<   y  =  x  +  f  +  f  +  tjr  .... 
2.   y  =  x-2x2  +  3x3-4:X4  +  ....  2       3      4 

4.   y  =  2x  +  5x2  +  8x3  +  llxi+  .... 

/y»  /V»2  /y»3  /y»4  /y,  /yvJ  ,y»3  /y»4 

5'   2/  =  2"4  +  6-8+  ""  6-  2/  =  |I+[3+[4  +  |5+-- 

7.   2/=2o;-4^+6a)5-8^+..,        8.    y  =  |  +  J  +  f  +  f  +  "" 

XII.   PERMUTATIONS  AND  COMBINATIONS 

275.  The  different  orders  in  which  things  can  be  arranged 
are  called  their  Permutations. 

Thus,  the  permutations  of  the  letters  a,  b,  c,  taken  two  at  a 
time,  are  ab,  ac,  ba,  be,  ca,  cb\  and  their  permutations,  taken 
three  at  a  time,  are  abc,  acb,  bac,  bca,  cab,  cba. 

276.  The  Combinations  of  things  are  the  different  collections 
which  can  be  formed  from  them  without  regard  to  the  order 
in  which  they  are  placed. 

Thus,  the  combinations  of  the  letters  a,  b,  c,  taken  two  at  a 
time,  are  ab,  be,  ca ;  for  though  ab  and  ba  are  different  permu- 
tations, they  form  the  same  combination. 

277.  To  find  the  number  of  permutations  of  n  different  things 
taken  tivo  at  a  time. 

Consider  the  n  letters,  a,  b,  c,  •••. 

In  making  any  particular  permutation  of  two  letters,  the 
first  letter  may  be  any  one  of  the  n ;  that  is,  the  first  place  can 
be  filled  in  n  different  ways. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  n  —  1  letters. 

Then,  the  whole  number  of  permutations  of  the  letfers  taken 
two  at  a  time  is  n(n  —  1). 

We  will  now  consider  the  general  case. 


204  ALGEBRA 

278.  To  find  the  number  of  permutations  of  n  different  things 
taken  r  at  a  time. 

Consider  the  n  letters  a,  b,  c,  •••. 

In  making  any  particular  permutation  of  r  letters,  the  first 
letter  may  be  any  one  of  the  n. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  n  —  1  letters. 

After  the  second  place  has  been  filled,  the  third  place  can  be 
filled  in  n  —  2  different  ways. 

Continuing  in  this  way,  the  rth  place  can  be  filled  in 
n  —  (r  —  1),  or  n  —  r  + 1  different  ways. 

Then,  the  whole  number  of  permutations  of  the  letters  taken 
r  at  a  time  is  given  by  the  formula 

nPr  =  n(n-l)(n-2)-..(n-r  +  l).  (1) 

The  number  of  permutations  of  n  different  things  taken  r  at  a  time  is 
usually  denoted  J)y  the  symbol  nPr. 

279.  If  all  the  letters  are  taken,  r  =  n,  and  (1)  becomes 

nPn  =  n(n-l)(n-2):-3-2-l  =  \n1_  (2) 

Hence,  the  number  of  permutations  of  n  different  things 
taken  n  at  a  time  equals  the  product  of  the  natural  num- 
bers from  1  to  n  inclusive.     (See  note,  §  181.) 

280.  To  find  the  number  of  combinations  of  n  different  things 
taken  r  at  a  time. 

The  number  of  permutations  of  n  different  things  taken  r  at 

a  time  is       n(n-l)(n~2)  ...  (n-r-f  1)  (§  278). 

But,  by  §  279,  each  combination  of  r  different  things  may 
have  \r  permutations. 

•  Hence,  the  number  of  combinations  of  n  different  things  taken 
rata  time  equals  the  number  of  permutations  divided  by  [r. 

That  is,  nCr  =  n(n-l)(n-2)..-(n-r  +  l) ;  Q]) 

\r 
The  number  of  combinations  of  n  different  things  taken  r  at  a  time  is 
usually  denoted  by  the  symbol  „Or. 


PERMUTATIONS   AND   COMBINATIONS  205 

281 .  Multiplying  both  terms  of  the  fraction  (3)  by  the  prod- 
uct of  the  natural  numbers  from  1  to  n  —  r  inclusive,  we  have 

C  =n(n-l)  ...(n-r  +  l)-(w-r)  -2-l_       b       ■ 
[r  X  1  •  2  •  •  •  (n  —  r)  \r_  \n  —  r 

which  is  another  form  of  the  result. 

282.  The  number  of  combinations  of  n  different  things 
taken  r  at  a  time  equals  the  number  of  combinations 
taken  n  —  r  at  a  time. 

For,  for  every  selection  of  r  things  out  of  n,  we  leave  a  selec- 
tion of  n  —  r  things. 

The  theorem  may  also  be  proved  by  substituting  n  —  r  f or  r,  in  the 
result  of  §  281. 

283.  Examples. 

i.  How  many  changes  can  be  rung  with  10  bells,  taking  7  at 
a  time  ? 

Putting  n  =  10,  r  =  7,  in  (1),  §  278, 

10P7  =  10-  9.  8-  7-  6.  5.  4  =  604800. 

2.  How  many  different  combinations  can  be  formed  with  16 
letters,  taking  12  at  a  time  ? 

By  §  282,  the  number  of  combinations  of  16  different  things,  taken  12 
at  a  time,  equals  the  number  of  combinations  of  16  different  things,  taken 
4  at  a  time. 

Putting  n  =  16,  r  =  4,  in  (3),  §  280, 

16.15.14.13  =  182(X 
16  1.2.3-4 

3.  How  many  different  words,  each  consisting  of  4  consonants 
and  2  vowels,  can  be  formed  from  8  consonants  and  4  vowels  ? 

The  number  of  combinations  of  the  8  consonants,  taken  4  at^a  time,  is 

8-7'°-5,or70. 
1- 2-3-4 


206  ALGEBRA 

The  number  of  combinations  of  the  4  vowels,  taken  2  at  a  time,  is 

i^3,or6. 
1-2' 

Any  one  of  the  70  sets  of  consonants  may  be  associated  with  any  one 
of  the  6  sets  of  vowels ;  hence,  there  are  in  all  70  x  6,  or  420  sets,  each 
containing  4  consonants  and  2  vowels. 

But  each  set  of  6  letters  may  have  ]_6,  or  720  different  permutations 
(§  279). 

Therefore,  the  whole  number  of  different  words  is 

420  x  720,  or  302400. 


EXERCISE   70 

i.  How  many  different  permutations  can  be  formed  with 
14  letters,  taken  6  at  a  time  ? 

2.  In  how  many  different  orders  can  the  letters  in  the  word 
triangle  be  written,  taken  all  together  ? 

3.  How  many  combinations  can  be  formed  with  15  things, 
taken  5  at  a  time  ? 

4.  A  certain  play  has  5  parts,  to  be  taken  by  a  company  of 
12  persons.  In  how  many  different  ways  can  they  be 
assigned  ? 

5.  How  many  combinations  can  be  formed  with  17  things, 
taken  11  at  a  time  ? 

6.  How  many  different  numbers,  of  6  different  figures  each, 
can  be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  each 
number  begins  with  1,  and  ends  with  9  ? 

7.  How  many  even  numbers,  of  5  different  figures  each,  can 
be  formed  from  the  digits  4,  5,  6,  7,  8  ? 

8.  How  many  different  words,  of  8  different  letters  each, 
can  be  formed  from  the  letters  in  the  word  ploughed,  if  the 
third  letter  is  o,  the  fourth  w,  and  the  seventh  e  ? 


PERMUTATIONS   AND   COMBINATIONS  207 

9.  How  many  different  committees,  of  8  persons  each,  can 
be  formed  from  a  corporation  of  14  persons  ?  In  how  many 
will  any  particular  individual  be  found? 

10.  There  are  11  points  in  a  plane,  no  3  in  the  same  straight 
line.  How  many  different  quadrilaterals  can  be  formed,  having 
4  of  the  points  for  vertices  ? 

11.  From  a  pack  of  52  cards,  how  many  different  hands  of 
6  cards  each  can  be  dealt  ? 

12.  A  and  B  are  in  a  company  of  48  men.  If  the  company 
is  divided  into  equal  squads  of  6,  in  how  many  of  them  will  A 
and  B  be  in  the  same  squad  ? 

13.  How  many  different  words,  each  having  5  consonants 
and  1  vowel,  can  be  formed  from  13  consonants  and  4  vowels  ? 

14.  Out  of  10  soldiers  and  15  sailors,  how  many  different 
parties  can  be  formed,  each  consisting  of  3  soldiers  and  3 
sailors  ? 

15.  A  man  has  22  friends,  of  whom  14  are  males.  In  how 
many  ways  can  he  invite  16  guests  from  them,  so  that  10  may 
be  males  ? 

16.  From  3  sergeants,  8  corporals,  and  16  privates,  how  many 
different  parties  can  be  formed,  each  consisting  of  1  sergeant, 
2  corporals,  and  5  privates  ? 

17.  Out  of  3  capitals,  6  consonants,  and  4  vowels,  how  many 
different  words  of  6  letters  each  can  be  formed,  each  beginning 
with  a  capital,  and  having  3  consonants  and  2  vowels  ? 

18.  How  many  different  words  of  8  letters  each  can  be 
formed  from  8  letters,  if  4  of  the  letters  cannot  be  separated  ? 
How  many  if  these  4  can  only  be  in  one  order  ? 

19.  How  many  different  numbers,  of  7  figures  each,  can  be 
formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  the  first,  fourth, 
and  last  digits  are  odd  numbers  ? 


208  ALGEBRA 

284.  To  find  the  number  of  permutations  of  n  things  which  are 
not  all  different,  taken  all  together. 

Let  there  be  n  letters,  of  which  p  are  a's,  q  are  &'s,  and  r  are 
c's,  the  rest  being  all  different. 

Let  N  denote  the  number  of  permutations  of  these  letters 
taken  all  together. 

Suppose  that,  in  any  particular  permutation  of  the  n  letters, 
the  p  a's  were  replaced  by  p  new  letters,  differing  from  each 
other  and  also  from  the  remaining  letters. 

Then,  by  simply  altering  the  order  of  these  p  letters  among 
themselves,  without  changing  the  positions  of  any  of  the  other 
letters,  we  could  from  the  original  permutation  form  [p  differ- 
ent permutations  (§  279). 

If  this  were  done  in  the  case  of  each  of  the  JV  original  per- 
mutations, the  whole  number  of  permutations  would  be  N  x  \p. 

Again,  if  in  any  one  of  the  latter  the  q  6's  were  replaced  by 
q  new  letters,  differing  from  each  other  and  from  the  remain- 
ing letters,  then  by  altering  the  order  of  these  q  letters  among 
themselves,  we  could  from  the  original  permutation  form  \q 
different  permutations ;  and  if  this  were  done  in  the  case  of 
each  of  the  N  x  \p  permutations,  the  whole  number  of  permu- 
tations would  be  N  x  ]/)  x  |  q. 

In  like  manner,  if  in  each  of  the  latter  the  r  c's  were  re- 
placed by  r  new  letters,  differing  from  each  other  and  from 
the  remaining  letters,  and  these  r  letters  were  permuted 
among  themselves,  the  whole  number  of  permutations  would  be 
Nx  [px  )_7  X  ]r. 

We  now  have  the'  original  n  letters  replaced  by  n  different 
letters. 

But  the  number  of  permutations  of  n  different  things  taken 
n  at  a  time  is  I  n  (§  279). 


Therefore,  N  x  \p  X  I  q  X  I  r  =  I  n  :  or,  N=  — =~-  • 

L       L_      L      1_  \p\q\r 

Any  other  case  can  be  treated  in  a  similar  maimer. 


PERMUTATIONS   AND   COMBINATIONS  209 

Ex.    How  many  permutations  can  be  formed  from  the  let- 
ters in  the  word  Tennessee,  taken  all  together? 

Here  there  are  4  e's,  2  n's,  2  s's,  and  1 1. 

Putting  in  the  above  formula  n  =  9,  p  =  4,  q  =  2,  r  =  2,  we  have 

l&      _5.6.7.8.9  =  378a 


L4(_2|2  2'2 

EXERCISE   71 

i.  In  how  many  different  orders  can  the  letters  of  the  word 
denomination  be  written  ?• 

2.  There  are  4  white  billiard  balls  exactly  alike,  and  3  red 
balls,  also  alike;  in  how  many  different  orders  can  they  be 
arranged  ? 

3.  In  how  many  different  orders  can  the  letters  of  the  word 
independence  be  written  ? 

4.  How  many  different  signals  can  be  made  with  7  flags,  of 
which  2  are  blue,  3  red,  and  2  white,  if  all  are  hoisted  for  each 
signal  ? 

5.  How  many  different  numbers  of  8  digits  can  be  formed 
from  the  digits  4,  4,  3,  3,  3,  2,  2,  1  ? 

6.  In  how  many  different  ways  can  2  dimes,  3  quarters,  4 
halves,  and  5  dollars  be  distributed  among  14  persons,  so  that 
each  may  receive  a  coin? 

285.  To  find  for  what  value  of  r  the  number  of  combinations 
of  n  different  things  taken  r  at  a  time  is  greatest. 

By  §  280,  the  mimber  of  combinations  of  n  different  things, 
taken  r  at  a  time,  is 

r  _n(tt-l)-(w-r+2)(n-r  +  l)  m 

*  '  L2.3...(r-l)r  '  W 


210  \i.(,i:i;i;  \ 

Also,  the  number  of  combinations  of  n  different  things,  taken 

r  —  1  at   I  t  line,  is 

»(n       \)  ■■■  \,t      (r      hill  n(n       I)  ., 

The  expression  (1)  Is  obtained  by  multiplying  the  expres- 
Bion  (2)  by     ■ — ,  or — i. 

The  latter  expression  decreases  as  r  increases. 
If.  then,  we  Bud  the  values  of  (I)  corresponding  to  the  Val- 
ues   L|   2j  •  ».  ■•••   Of   /*.  the   results   will    OOntinnally    increase   so 

i  «       /•  !    1    .    '     -, 

Long  as  is  >  l. 

/' 

I.  Suppose  a  even;  and  let  n    »2m,  where  mn  ia  a  positive 
integer. 

Then,  ™ — iX-  beoomes  — ■ - — 

r  r 

If,      w>    2w'-r  +  1    becomes   '"  j   1,and  is        1. 

2  m  —  r +1 .  i  ,",       i 

If  r-m  +  1,  -         ;.  becomes  r  and  is  -     I. 

Then,  ,0!  will  bave  its  greatest  value  when  r»HM 

II.  Suppose  N  Oddj  and   let   n     ■  %2m   I    1.  where  m  is  a  posi- 
tive integer, 

1  hen.  becomes  " ' 

r  r 

if  r     ».  -  -     -  j  "  beoomes  "      -j  and  is  >l, 

r  m 

[fr«m+l,  -  -  beoomes  -  4~ti  and  equals  1, 

r  »<  4- 1 

2  m     r  I  4>                   tw 
If  r     in  I  2,  -  -  beoomes ^,  and  is  <1. 

r 


DETERMINANTS 


211 


Then,  nCr  will  have  its  greatest  value  when  r  equals  in  or 


1        n 

—  or  — 


+  1. 


m  -f- 1 ;  that  is, 


Then,  n(7r  will  have  its  greatest  value  when  r  equals 

n  4- 1 

or  — ~— ;  the  results  being  the  same  in  these  two  cases. 

Li 

XIII.    DETERMINANTS 
286.    The  solution  of  the  equations 
|  aYx  +  b^y  =  clf 
\  a&  -f-  bgj  =  c2, 

is  x  —    b&  —  b&      y  =  °2al  ~  Cia2  . 

afii  —  ajbi  aYb2  —  a2&i 

The  common  denominator  may  be  written  in  the  form 
al9     h 


-1 


(1) 


This  is  understood  as  signifying  the  product  of  the  upper 
left-hand  and  lower  right-hand  numbers,  minus  the  product  of 
the  lower  left-hand  and  upper  right-hand. 

The  expression  (1)  is  called  a  Determinant  of  the  Second  Order. 

The  numerators  of  the  above  fractions  can  also  be  expressed  as  deter- 
minants ;  thus, 

b-zCi  —  b\C2  = 


Cl, 

bi 

,  and  c2a\  - 

-  Ci<Z2  = 

«i» 

Cl 

C2» 

bf 

«2, 

c2 

287.   The  solution  of  the  equations 

axx  +  bjy  +  cxz  =  dly 

a^x  -\-b2y  +  c2z  =  d2, 

.asx  +  bsy  +  csz  =  d5y 

,g  ^  =  dfi&a  —  dfifo  +  d2b^  —  d2blcs  +  dsbYc2  —  d3bfr  t         ~ . 

afyfy  —  aj>&  +  aj)si\  —  dib^s  +  ajbxc2  —  ajbfr 

with  results  of  similar  form  for  y  and  z. 


212 


ALGEBRA 


The  denominator  of  (1)  may  be  written  in  the  form 

a2,     b2,     c2  .  (2) 

This  is  understood  as  signifying  the  sum  of  the  products  of  the 

numbers  connected  by  lines 
parallel  to  a  line  joining  the 
upper  left-hand  corner  to  the 
lower  right-hand,  in  the  fol- 
lowing diagram,  minus  the 
sum  of  the  products  of  the 
numbers  connected  by  lines 
parallel  to  a  line  joining 
the  lower  left-hand  corner 
to  the  upper  right-hand. 

The  expression  (2)  is  called  a  Determinant  of  the  Third  Order. 

The  numerator  of  (1)  can  also  be  expressed  as  a  determinant,  as  follows : 

ds,     b3,     c8 
as  may  be  verified  by  expanding  it  by  the  above  rule. 


EXERCISE   72 


Evaluate  the  following: 


2. 


14    15 

9    12 

2  x  —  ?/     2  x  -f-  y 

2x  +  y 

2x-y 

4     3 


2    1 

8     7 


4- 

3    4    5 
9    12 

6     7     8 

• 

5- 

15     12 
8      4 
6      5 

2 

4 

6. 


12    9 

0     2 
5     4 


DETERMINANTS 


213 


Show  that 

6  8  10 

7  5  10 

11  6  10 

8   5  4 

12   9  6 

. 

14  16  7 

Show  that 

7-4    8 

3.   6  -9 

8-5   1 

3 

=  6 


10 
10 


10 
10 


+  11 


10 
10 


-4  8 

7  8 

7 

-4 

3 

-5  13 

+  6 

8  13 

+  9 

8 

-5 

It  is  found  in  geometry  that  if  the  vertices  of  a  triangle  are 
at  the  points  x  =  2,  ?/  =  3 ;  a?  =  4,  y  =  5;  x  =—  1,  y  =  4,  the 
area  of  the  triangle  is  found  to  be 

2     3     1 

,  the  abscissas  (§  46)  forming  the  first 


Area  : 


4     5     1 
-14     1 

column  of  the  determinant,  the  ordinates  the  second  column, 
and  the  third  column  being  l's. 

Find  the  areas  of  the  triangles  whose  vertices  are  at  the 
following  points: 

io.*   x=—  2,  y=l\  x=£,  y=l;  x=2,  y=6.     Make  diagram. 

ii.  x  =  —  4,  2/ =  3;  a  =  4,  ?/ =  3 ;  x  =  2,y  =  —7.  Make 
diagram. 

12.  x  =  2,  2/  =  4;  x  =  8,  y  =  A-,  x=—  1,  y  =  —  9.  Make 
diagram. 

13.  x  =  —  5,  y  =  S-y  x  =  5,  y  =  3-,  x  =  0,  y=—3.  Make 
diagram. 

14.  a=-2,  y  =  8;  a  =  -2,  y=-2;  #  =  5,  2/=~4- 
Make  diagram. 

*  If  your  area  is  negative,  its  absolute  value  is  the  area  sought.  A 
change  in  the  order  of  selecting  the  vertices  will  change  the  sign  of  the 
area,  but  not  the  absolute  value  of  the  area. 


214 


ALGEBRA 


288.  The  numbers  in  the  first,  second,  etc.,  horizontal  lines 
of  a  determinant  are  said  to  be  in  the  first,  second,  etc.,  rows, 
respectively  ;  and  the  numbers  in  the  first,  second,  etc.,  vertical 
columns,  in  the  first,  second,  etc.,  columns. 

The  numbers  constituting  the  determinant  are  called  its 
elements,  and  the  products  in  the  expanded  form  its  terms. 

Thus,  in  the  determinant  (2),  of  §  287,  the  elements  are  <i\,  a2,  a3, 
etc.,  and  the  terms  aib2C3,  —  aib^,  etc. 


289.  If,  in  any  permutation  of  the  numbers  1,  2,  3,  •••,  n, 
a  greater  number  precedes  a  less,  there  is  said  to  be  an 
inversion. 

Thus,  in  the  case  of  five  numbers,  the  permutation  4,  3,  1,  5,  2  has  six 
inversions  ;  4  before  1,  3  before  1,  4  before  2,  o  before  2,  5  before  2,  and 
4  before  3. 


290.    General  Definition  of  a  Determinant. 
Consider  the  n2  elements 


a9 


xn,  3? 


(1) 


The  notation  in  regard  to  suffixes,  in  (1),  is  that  the  first 
suffix  denotes  the  row,  and  the  second  the  column,  in  which 
the  element  is  situated. 

Thus,  akt1.  is  the  element  in  the  &th  row  and  rth  column. 

Let  all  possible  products  of  the  elements  taken  n  at  a  time 
be  formed,  subject  to  the  restriction  that  each  product  shall 
contain  one  and  only  one  element  from  each  row,  and  one  and 
only  one  from  each  column,  and  write  them  so  that  the  first 
suffixes  shall  be  in  the  order  1,  2,  3,  ...,  n. 

This  is  equivalent  to  writing  all  the  permutations  of  the  numbers  1,  2, 
3,  •••,  n  in  the  second  suffixes. 


DETERMINANTS 


215 


Make  each  product  4-  or  —  according  as  the  number  of  in- 
versions in  the  second  suffixes  is  even  or  odd. 

The  expression  (1)  is  called  a  Determinant  of  the  nth 
Order. 

The  number  of  terms  in  the  expanded  form  of  a  determinant  of  the  nth 
order  is  [n  (§  279). 

291.  The  elements  lying  in  the  diagonal  joining  the  upper 
left-hand  to  the  lower  right-hand  corner,  of  a  determinant,  are 
said  to  be  in  the  princijial  diagonal;  the  term  whose  factors  are 
the  elements  in  the  principal  diagonal  is  always  positive. 


P 


292.  It  may  be  shown  that  the  definition  of  §  290  agrees 
with  that  of  §  287. 

For  consider  the  determinant 

al,  1>  a\,  2)  a\,  3 
a2,  1?  ®>2,  2?  ^2,  3 
%,  1)        ^3,  2?        #3,  3 

The  products  of  the  elements  taken  three  at  a  time,  subject  to  the 
restriction  that  each  product  shall  contain  one  and  only  one  element 
from  each  row,  and  one  and  only  one  from  each  column,  the  first 
suffixes  being  written  in  the  order  1,  2,  3,  are 

0\,  1  #2,  2  #3,  3?    #1,  1  #2,  3  «3,  2,    (L\,  2  0>2,  1  #3,  3?    #1,  2  #2,  3  #3,  1?    #1,  3  #2,  1  #3,  2> 

and  «i,3  a2, 2  #3,  i- 

In  the  first  of  these  there  are  no  inversions  in  the  second  suffixes ;  in 
the  second  there  is  one,  3  before  2  ;  in  the  third  there  is  one  ;  in  the 
fourth,  two  ;  in  the  fifth,  two  ;  in  the  sixth,  three. 

Then  by  the  rule  of  §  290,  the  first,  fourth,  and  fifth  products  are 
positive,  and  the  second,  third,  and  sixth  are  negative ;  and  the  ex- 
panded form  is 


0\,  1  0,2,  2  053,  3  —  &1,  1  0,2,  3  #3,  2  —  #1,  2  #2,  1  0%,  3  +  Q>\,  2  &2,  3  <fe,  1 
+  «1,  3  02,  1  O3,  2  ~  «1,  3  02,  2  «3,  U 

which  agrees  with  §  287. 


216 


ALGEBRA 


293.  The  expanded  form  of  the  determinant  (1),  §  290,  may 
also  be  obtained  by  writing  the  second  suffixes  in  the  order 
1,  2,  3,  •••,  n,  and  making  each  product  -for  -  according 
as  the  number  of  inversions  in  the  first  suffixes  is  even  or 
odd. 

For  let  the  absolute  value  of  any  term,  obtained  by  the  rule  of  §  290,  be 
«i,p  «2,«  •••  an,r;  (1) 

where p,  q,  •••,  r  is  a  permutation  of  1,  2,  ••-,  n. 
This  is  obtained  from  the  first  term 

«1,1  <*2,2  -•  «»,n  (2) 

by  changing  second  suffixes,  1  to  p,  2  to  g,  • 
Since  j»,  q,  •  ••,  r  is  a  permutation  of  1,  2, 


,  n  to  r. 

.,  ?i,  (2)  may  be  written 


...  ar, 


and  (1)  may  be  obtained  from  this  by  changing  first  suffixes,  p  to  1,  q  to 
2,  ••.,  r  to  n. 

In  these  two  ways,  we  have  the  same  number  of  interchanges 
of  two  suffixes,  and  hence  the  term  (1)  will  have  the  same  sign. 

PROPERTIES   OF   DETERMINANTS 

294.    A  determinant  is  not  altered  in  value  if  its  rows 
are  changed  to  columns,  and  its  columns  to  rows. 

Consider  the  determinants 


«i,i> 


*2,  1> 


^2,  2? 


H%  n 


-%  n 


xn,  1) 


and 


ai,  2? 


tt»,  1 
««,2 


l2,  nj 


Since  the  second  suffixes  of  the  first  determinant  arc  tlic 
same  as  the  first  suffixes  of  the  second,  if  the  first  determinant 
be  expanded  by  the  rule  of  §  290,  and  the  second  by  the  rule 
of  §  293,  the  results  will  be  the  same. 

Therefore  the  determinants  are  equal. 


DETERMINANTS 


217 


295.  A  determinant  is  changed  in  sign  if  any  two  con- 
secutive rows,  or  any  two  consecutive  columns,  are 
interchanged. 

Consider  the  determinants 


a     b     c 

b     a 

c 

d     e    f 

and 

e     d 

f 

g     h     Jc 

h     g 

k 

Evaluating  each  determinant  as  in  Exercise  72,  we  have 
aek  +  dhc  +  bfg  —  gee  —  bdk  —  fha 
and  bdk  -f  egc  +  afh  —  hdc  —  aek  —  fgb, 

each  term  of  the  second  determinant  being  the  negative  of  the  correspond- 
ing term  in  the  first.  The  second  determinant  has  therefore  the  same 
absolute  value  as  the  first,  but  of  opposite  sign.  In  a  manner  similar  to 
that  used  in  §  294,  it  may  be  shown  that  this  property  holds  for  determin- 
ants of  the  nth  order. 

It  follows,  from  §§  294  and  295,  that  if  two  consecutive 
rows  are  interchanged,  the  sign  of  the  determinant  is  changed. 

296.  A  determinant  is  changed  in  sign  if  any  two 
rows,  or  any  two  columns,  are  interchanged. 

Consider  the  m  letters  a,  b,  c,  •••,  e,  f,  g. 

By  interchanging  a  with  &,  then  a  with  c,  and  so  on  in  succession  with 
each  of  the  m  —  1  letters  to  the  right  of  a,  a  may  be  brought  to  the  right 
of  g. 

Then,  by  interchanging  g  with  /,  then  g  with  e,  and  so  on  in  succession 
with  each  of  the  m  —  2  letters  to  the  left  of  g,  g  may  be  brought  to  the 
left  of  b. 

That  is,  a  and  g  may  be  interchanged  by  (m  —  1)  +  (ni  —  2),  or 
2  m  —  3,  interchanges  of  consecutive  letters ;  that  is,  by  an  odd  number 
of  interchanges  of  consecutive  letters. 

It  follows  from  the  above  that  any  two  rows,  or  any  two  columns,  of 
a  determinant  may  be  interchanged  by  an  odd  number  of  interchanges 
of  consecutive  rows  or  columns. 

But  every  interchange  of  two  consecutive  rows  or  columns  changes 
the  sign  of  the  determinant  (§  295).  ** 

Therefore  the  sign  of  the  determinant  is  changed  if  any  two 
rows,  or  any  two  columns,  are  interchanged. 


218 


ALGEBRA 


297.    Cyclical  interchange  of  rows  or  columns. 

By  n  —  1  successive  interchanges  of  two  consecutive  rows,  the 
first  row  of  a  determinant  of  the  nth  order  may  be  made  the  last. 
Thus,  by  §  295,  the  determinant 


a2,     b< 


is  equal  to  (—  l)n 


an, 


The  above  is  called  a  cyclical  interchange  of  rows. 

In  like  manner,  by  n  —  1  successive  interchanges  of  two  con- 
secutive columns,  the  first  column  of  a  determinant  of  the  nth 
order  may  be  made  the  last. 

298.  If  two  rows,  or  two  columns,  of  a  determinant 
are  identical,  the  value  of  the  determinant  is  zero. 

Let  D  be  the  value  of  a  determinant  having  two  rows,  or  two  columns, 
identical. 

If  these  rows,  or  columns,  are  interchanged,  the  value  of  the  resulting 
determinant  is  -  D  (§  296). 

But  since  the  rows,  or  columns,  which  are  interchanged  are  identical, 
the  two  determinants  are  of  equal  value. 

Hence,  D  —  —  D  ;  and  this  equation  cannot  be  satisfied  by  any  value 
of  D  except  0. 

Ex.     Evaluate  the  following  determinant : 
a     a     d 

=  abk  4-  bed  4-  cea  —  cbd  —  abk  —  ace. 

=  0. 


299.  If  each  element  in  one  column,  or  in  one  row,  is 
the  sum  of  m  terms,  the  determinant  can  be  expressed 
as  the  sum  of  m  determinants. 

Consider  the  determinant 


a2,  1> 


a2,  r) 


(1) 


<ln 


a 


n,  r) 


DETERMINANTS 


219 


Let  each  element  in  the  rth  column  be  the  sum  of  m  terms,  as  follows: 
ahr=  hi  +  ci+  ••■'•  +/i, 

«2,r  =  fo  +  C2  +    '••    +/2, 


Let  «i,?, 
then       «i, 


a, 


an,r=  K  +  Cn+    •••    +/„. 

g,r  •••  «n,«  be  the  absolute  value  of  one  of  the  terms  of  (1); 


(in 


(in, 


(ihp  •••  (bq  +  cq=  •••  +  /g)  ...  a„,8 

It  is  evident  from  this  that  the  determinant  (1)  can  be  expressed  as 
the  sum  of  the  determinants 


«2,  1, 


hi 


#1,  n 

<t2,n 


bn, 


On, 


+  "•   + 


01,  I1 


fu 

/11 


#1,n 
«2,n 


«n,  1, 


Jni 


a»,n 


300.  If  all  the  elements  in  one  column,  or  in  one  row, 
are  multiplied  by  the  same  number,  the  determinant  is 
multiplied  by  this  number. 

Consider  the  determinant 


a2,  i> 


; 


"1,  r> 

a2,  rl 


<Xo 


an,ly        '">       8*»rj        ""        an, 

Multiplying  each  element  in  the  rth  column  by  m,  we  have 
«i,i,     •••,     mai,n     *-i     «i, 


«2,1, 


ma2,r, 


«2, 


(i) 


(2) 


Let  «iiP  •••  ag,  r  •••  an,s  be  the  absolute  value  of  one  of  the  terms  of  (1). 

Replacing  aq,  r  by  maq> ,.,  the  absolute  value  of  the  corresponding  term 
of  (2)  is  mai,p  •••  aq>r  •••  «»,«• 

It  is  evident  from  this  that  the  determinant  (2)  equals  m  times  the 
determinant  (1). 

Ex.     Consider  the  values  of 

9 

h  . 
k 

•  Evaluating,       m(aek  -f-  bfg  +  chd  —  ceg  —  dbk  —  afh)        and 
maek  +  mbfg  +  mchd  —  mceg  —  dmbk  —mafli,  which  are  identical. 


a 

d     g 

ma 

d 

b 

e     h 

and 

mb 

e 

c 

f    * 

mc 

f 

220 


ALGEBRA 


301 .  If  all  the  elements  in  any  column,  or  row,  be  mul- 
tiplied by  the  same  number,  and  either  added  to,  or  sub- 
tracted from,  the  corresponding  elements  in  another 
column,  or  row,  the  value  of  the  determinant  is  not 
changed. 

Let  the  elements  in  the  rth  column  of  the  following  deter- 
minant be  multiplied  by  m,  and  added  to  the  corresponding 
elements  in  the  qth  column. 


«1>        •'•■>        ®q, 

'"t 

Ctr, 

t       &n 

h,    •>•$    bq,     •••, 

br, 

,       &n 

(1 

#i?     •••?     1cq<)     •••? 

fori 

,       fCn 

We  then  obtain  the  determinant 

0>U        '"•)       ®q  +   mar, 

-•» 

Or, 

'**! 

&n 

6i,     ••,     bq  +  mhr, 

'"5 

&n 

-' 

bn 

(2 

ki,     •  •,     kq  +  mkr, 

-.., 

fori 

***? 

K 

which  by  §§  299  and  300,  is  equal  to 

Q>ll       *"%       Qqi       "'$       Qri        '"i       Q>n 

«1, 

'"  i 

«r, 

,     an     •• 

',      «n 

b\,     —,     bqj     •••,     &PJ     •••,     bn 

+  m 

hi 

br, 
rCr, 

5r,     •• 

*»      tin 

Kit        '*•$        Kq-j 

..? 

AV,       "**}      "'W 

But  the  coefficient  of  m  is  zero  (§  298). 
Whence,  the  determinant  (2)  is  equal  to  (1). 

302.    Minors. 

If  the  elements  in  any  m  rows  and  any  m  columns  of  a 
determinant  of  the  r*th  order  be  erased,  the  remaining  elements 
form  a  determinant  of  the  (n  —  m)th  order. 

This  determinant  is  called  an  mth  Minor  of  the  given  deter- 
minant.                                                               h      „  n      „ 

<l\,      Oi,      Ci,  «i,       €i 

ao,       bo,       Co,  0*2,       «2 

«3,       ftfcl       C3,  0*3,       €3 

«4,       &4,       C4,  #%,       €4 

as,      &5,      C6,  d6,      <5 


Thus, 


«i, 

di, 

ei 

«3, 

0*3, 

€3 

«5, 

a*5, 

€5 

is  a  second  minor  of 


DETERMINANTS 


221 


obtained  by  erasing  the  second  and  fourth  rows,  and  the  second  and 
third  columns. 

303.    To  find  the  coefficient  of  alt  x  in  the  determinant 

^2,1?       tt2,  2>        ***j       ^2,  n 


(1) 


By  §  290,  the  absolute  values  of  the  terms  which  involve 
ah  x  are  obtained  by  forming  all  possible  products  of  the  ele- 
ments taken  n  at  a  time,  subject  to  the  restrictions  that  the 
first  elements  shall  be  ah  x  and  that  each  product  shall  contain 
one  and  only  one  element  from  each  row  except  the  first,  and 
one  and  only  one  from  each  column  except  the  first. 

It  is  evident  from  this  that  the  coefficient  of  ah  j  in  (1)  may 
be  obtained  by  forming  all  possible  products  of  the  following 
elements  taken  n  —  1  at  a  time, 

^2,2?       ^2,3?        '">       ®2,n 
a3,  2?        a3, 3>        ***?        a3,  n 


subject  to  the  restriction  that  each  product  shall  contain  one 
and  only  one  element  from  each  row,  and  one  and  only  one  from 
each  column,  writing  the  first  suffixes  in  the  order  2,  3,  •  •,  n, 
and  making  each  product  +  or  —  according  as  the  number  of 
inversions  in  the  second  suffixes  is  even  or  odd. 
Then  by  §  290,  the  coefficient  of  alt  x  is 

Ojo  o j        ^2  3?         *  *  *?        ^2  n 
tt3, 2)        a3, 3>        "'J        a3,n 


that  is,  the  minor  obtained  by  erasing  the  first  row  and  the 
first  column  of  the  given  determinant. 


222 


ALGEBRA 


304.  By  aid  of  §  303,  a  determinant  of  any  order  may  be 
expressed  as  a  determinant  of  any  higher  order. 

1,    0,     0,     0,     0 


Thus, 


1, 

o, 

o, 

0 

a%% 

hi 

Cl 

()7 

Cll, 

hi, 

Cl 

«2, 

&2, 

c2 

= 

0, 

«2? 

&2, 

c2 

0*, 

6»? 

c3 

0, 

az, 

&•, 

c3 

0,  1,  0,  0,  0 

0,  0,  GL\,  b\,  C\ 

0,  0,  a2,  b2,  c2 

0,  0,  az,  b3,  c3 


etc. 


305.   Coefficient  of  any  Element  of  a  Determinant. 
To  find  the  coefficient  of  b3,  in  the  determinant 


"■■i, 


K 
bo, 


CO 


(-1)3 


%>        ^3?       °3 

By  two  interchanges  of  consecutive  rows,  the  last  row  may  be  made 
the  first ;  thus,  by  §  295,  the  determinant  equals 

a3,     &3,     c3 
(-1)2  ai,     h,     ft  •  (2) 

a*L,       b2,       C2 

By  interchanging  the  first  two  columns,  the  determinant  (2)  equals 

b3,     az,    cs 
(- 1)3  6i,     oi,     4  •  (3) 

b2,    a2,    c2 
By  §  303,  the  coefficient  of  b3,  in  (3),  is 

a\,     ci 
a2,     c2 


That  is,  the  coefficient  of  the  element  in  the  third  row  and  second 
column  equals  (—  1)*+2  multiplied  by  that  minor  of  (1)  which  is  obtained 
by  erasing  the  third  row  and  second  column. 

We  will  now  consider  the  general  case  ;  to  find  the  coefficient  of  a*,r  in 
the  determinant 

a\,i,     •••,     <*i,r»     •••»     «i, 


ak,  i, 


an,u 


aic,r 


',     an,ri 


«*,* 


(4) 


DETERMINANTS 


223 


By  k  —  1  interchanges  of  consecutive  rows,  and  r  —  1  interchanges  of 
consecutive  columns,  the  element  a&,  r  may  be  brought  to  the  upper  left- 
hand  corner. 

Thus,  by  §  295,  the  determinant  equals 

Then,  by  §  303,  the  coefficient  of  aA,  r  is 


(_  1)*+r-2 


$n,  1»       ***?      ^n,  n 


But 


(_  l)*+r-2  -  (_  !)t+r  +(_  1)«  =  (_  1) 


fc+r 


Hence,  the  coefficient  of  the  element  in  the  ftth  row 
and  rth  column  equals  (—  l)*+r?  multiplied  by  the  minor  of 
(4)  which  is  obtained  by  erasing  the  Zcth  row  and  rth 
column. 


306.  By  aid  of  §  305,  a  determinant  of  any  order  may 
be  expressed  in  terms  of  determinants  of  any  lower 
order. 

Thus,  since  every  term  of  a  determinant  contains  one  and  only  one 
element  from  the  first  row,  we  have, 


au  &i,  ci,  di 

#2?  &2>  Cli  d'l 

«3,  63,  c3,  dj 

#4,  64,  C4,  C?4 


|&2» 

c2, 

$2 

«1    63, 

C3, 

<*3 

-h 

|&4i 

C4, 

^4 

«2»  C2,  <?2 
«3?  ^3>  ^3 
«4,       C4,       (?4 


«2> 

&2» 

<^2 

«3, 

fc. 

C*3 

rr* 

«4, 

64, 

<?4 

«2, 

&2, 

C2 

«3, 

fe 

c3 

«4, 

&4» 

C4 

+  Ci 


and  each  of  the  latter  determinants  may  in  turn  be  expressed  in  terms  of 
determinants  of  the  second  order, 


224 


ALGEBRA 


307.   Evaluation  of  Determinants. 

The  method  of  §  306  may  be  used  to  express  a  determinant 
of  any  order  higher  than  the  third  in  terms  of  determinants  of 
the  third  order,  which  may  be  evaluated  by  the  rule  of  §  290. 

The  theorem  of  §  301  may  often  be  employed  to  shorten  the 
process. 


i.   Evaluate 


Subtracting  the  first  row  from  the  last,  twice  the  first  row  from  the 
second,  and  three  times  the  first  row  from  the  third  (§  301),  the  determi- 
nant becomes 


5, 

7, 

8, 

6 

U, 

16, 

13, 

11 

14, 

24, 

20, 

23 

7, 

13, 

12, 

2 

5,  7,  8,  6 

1,  2,  -3,  -1 
-1,  3,  -4,  5 

2,  6,  4,  -4 


=  2 


5,  7,  8,  6 

1,  2,  -3,  -1 

-1,  •  3,  -4,  5 

1,  3,  2,  -2 


,  by  §  300 


Subtracting  five  times  the  second  row  from  the  first,  adding  the 
second  row  to  the  third,  and  subtracting  the  second  row  from  the  last, 
we  have 


0,  -3,  23,  11 

1,  2,  -3,  -1 
0,  5,  -7,  4 
0,  1,  5,  -1 


=  -2 


8, 

23, 

11 

5, 

-7, 

4 

1, 

5, 

-1 

,  by  §  305. 


The  object  of  the  above  process  is  to  put  the  given  determinant  in  such 
a  form  that  all  but  one  of  the  elements  in  one  column  shall  be  zero  ;  the 
determinant  can  then  be  expressed  as  a  determinant  of  the  third  order 
by  §  305. 

The  last  determinant  may  be  evaluated  by  §  287  ;  but  it  is  better  to 
subtract  five  times  the  first  column  from  the  second,  and  then  add  the 
first  column  to  the  last ;  thus, 


-2 


-3,  38,  8 
5,  -32,  9 
1,  0,     0 


=  -2 


38,     8 
-32,     9 


=  -  2  (342  +  256)  =  -  1196. 


DETERMINANTS 


225 


The  artifice  used  in  the  following  example  is  often  of  use  in 
evaluation  of  determinants : 


2.   Evaluate 


x" 

f 

z3 


If  x  be  put  equal  to  ?/,  the  determinant  has  two  rows  identical,  and 
equals  zero  (§  298). 

Then,  as  —  y  must  be  a  factor  (§  105) ;  and  in  like  manner,  y  —  z  and 
z—  x  are  factors. 

Let  the  given  determinant  =  X  (x  —  y)  (y  —  z)  (z  —  x). 

To  determine  X,  we  observe  that  as,  ?/,  and  z  are  factors  of  the  deter- 
minant ;  then,  Xmust  equal  xyz,  as  it  is  evident  by  noticing  that  the  first 
term  in  the  expanded  form  is  +  xy2zs,  and  the  value  of  the  determinant  is 
xyz{x  -y)(y  —  z)  {z  —  x). 


EXERCISE   73 

Evaluate  the  following : 
8,     24,     16 


3, 
21, 

12, 
14, 


30, 
41, 


16,    39, 
13,     14, 


14, 
15, 


15, 
13, 


15 

26 

6 

7 

8 

15 

13 

14 


4- 


5- 


a-f  by 

b+c, 
o  +a, 


y, 


a, 
h 
c> 

1 
1 
1 


7. 


x, 
■4, 


y> 
-T, 

6, 


6.      3, 

6, 

Plot  this  line  (§51). 
y  =  5,  and  as  =  6,  y 

4, 


2/, 
12, 


=  0. 


=  0. 


Are  the  points  x  —  —  4,  ?/  =  —  7, 
and  as  =  8,   ?/  =  6,    on    this    line? 


8. 


15,  12,  11, 
14,  -4,  -8, 

16,  10,  -2, 
10,  12,  3, 


6 
25 

5 
6 


9- 


io. 


Are  the  points  x  =  3, 
=  —  12,  on  this  line? 

0,         0,         3 


7, 

9, 

o, 

0 

8, 

o, 

o, 

10 

o, 

o, 

-6, 

-8 

o, 

3, 

5, 

9 

5, 

o, 

7, 

8- 

4, 

9, 

o, 

16 

8, 

16, 

4, 

0 

226 


ALGEBRA 


4,     0, 

12,     10 

8, 

t 

0,     0, 

0 

0,     3,       0,       0 

0,       6,     15, 

18 

II. 

10,     0,       6,       4 
24,     0,       3,     12 

12. 

0,     12,     98, 

0,       2,       5, 

104 
6 

5,    -4, 

0, 

15 

-4,      5, 

15,       0 

13.    Show  that 

12,        0, 
17,    -9, 

5, 
-9, 

4 
3 

= 

0,    12, 
-9,    17, 

4,       5 
3,  -9 

6,        8, 

8, 

-2 

8,      6, 

-2,       8 

14,      3, 

-13 

1 

-13,    21,- 

-17,  -11 

14.    Show  that 

8,-5, 
21,    24, 

21 

-17 

2 
1 

= 

14,      8, 

3,  -5, 

21,    -6 
24,    -9 

-6,  -9, 

-11 

p  2 

1,      2, 

1,        2 

5,     6, 

12, 

16 

5,  2, 

3,      16 

15.    Show  that 

7,     9,  - 
18,  27, 

-16, 
36; 

-15 
9   : 

=  108 

7,  3, 
2,  1, 

-4,  -15 
1,        1 

2,     9, - 

-20, 

-11 

2,3, 

-5,  -11 

16,     8, 

7, 

-4 

16.   Evaluate 

8,     6, 

7,     5, 

3, 
-11, 

2 
3 

-4,     7, 

8, 

0 

in,     y,     n, 

X 

17.    Evaluate 

x,    y,     n, 

«i   »»    y, 

m 
m 

m,    n, 

y, 

X 

It  is  shown  in  geometry  that  if  a  straight  line  passes  through 
the  points  x  =  xl9  y  =  yly  x  —  x2,  y  =  y2,  where  xx,  x2,  yly  y2  are 
definite  values  of  x  and  2/,  the  equation  of  the  line  is  found 
from  the  determinant 

1 

1   =0. 

1 


X 

y 

xx 

V\ 

x2 

2/2 

DETERMINANTS 


227 


Find  the  equations  of  the  lines  passing  through  the  following 
points : 

18.  as  =  1,  y  =  3;  #=—  2,  y  s4 

19.  x  =  0,  y  =  8 ;  x  =  8,  y  =  0. 

20.  #  =  —  1,  2/  =  2 ;  a?  =  3,  y  =  —  1. 

308.   Let  Ar,  Bn  •  ••,  iTr,  denote  the  coefficients  of  the  ele- 
ments a,.,  br,  •••,  kr,  respectively,  in  the  determinant. 


hi     ^1? 


Kq 


K, 


"n 


(1) 


Then,  since  every  term  of  the  determinant  contains  one  and 
only  one  element  from  the  first  column,  the  value  of  the 
determinant  is 

A1a1  +  A2a2-\-  •••  +  Anan. 

In  like  manner,  the  value  of  the  determinant  also  equals 

,      BA  +  B2b2  +  . ..  +  BHbn,  •  •  •,  KJh  +  KM2  +  •  • .  4-  &Jc* 

309.  If  mlf  m2,  •  ••,  mn  are  the  elements  in  any  column  of  the 
determinant  (1),  of  §  308,  except  the  first, 

A^in^  +  A2m2  -+-  •  •  •  -f  Anmn 

is  the  value  of  a  determinant,  which  differs  from  (1)  only  in 
having  m1?  m2,  •••,  mn  instead  of  alt  a2,  •••,  an  as  the  elements  in 
the  first  column. 

Then  A1m1  +  A2m2  -f  •••  +  Anmn  =  0 ; 

for  it  is  the  value  of  a  determinant  which  has  two  columns 
identical. 

In  like  manner,  if  m1?  ra2,  •••,  mn  are  the  elements  in  any 
column  of  the  determinant  (1),  except  the  second,  ^ 

Bim^  +  B2m2  +  •  •  •  +  5nmn  =  0 ; 
and  so  on. 


228 


ALGEBRA 


SOLUTION   OP   EQUATIONS 


310.    Let  it  be  required  to  solve  the  following  system  of  n 
linear,  simultaneous  equations,  involving  n  unknown  numbers : 


0&1  + 


+  q&r  + 


+  hxH  =  plm 


la„a1  +  •••  +  q,flr+  ■•■•  4-&A=2>„. 


(1) 

(2) 
(3) 


Let  Qu  Q2,  •  ••,  Qn  denote  the  coefficients  of  the  elements  #i,  q2,  •••,  qn 
respectively,  in  the  determinant 


D-- 


Q2, 


»j  k2 


•j  kn 


Multiplying  equations  (1),  (2),  ...,  (3)  by  §b  Q2,  •  •,  Qm  respectively, 
and  adding,  we  have 


xiiQiai  +  Q2a2  +  •••  +  Qnan)  +  ••• 
+  Xr{Qiqi  +  §2^2  +  •••  +  Qnqn)  +  •" 
+  xn(Qiki  +  Q2k2  +  -.  +  Qnkn)  =  QiPi  +  Q2p2  + 


+  §nPn. 


By  §  309  the  coefficient  of  each  unknown  number,  except  xr,  is  zero. 

By  §  308,  the  coefficient  of  xr  is  D ;  also,  the  second  member  is  the 
value  of  a  determinant  which  differs  from  D  only  in  having  pi,  p2,  •••»  Pn 
instead  of  qu  q2,  •••,  qn  as  the  elements  in  the  rth  column ;  denoting  the 
latter  by  Dn  we  have 


xrD  =  Dr,  and  xr  = 


I) 


311.   i£#.     Find  the  value  of  y  from  the  equations 

(3x-5y  +  7z  =  28. 
i2a  +  6</-9z=-23. 
[4:X-.2y-5z=        9. 

The  denominator  of  the  value  of  y  is  the  determinant 

3,-5,  7 
2,  6,-9 
4,     -2,     -5 


DETERMINANTS 


229 


The  numerator  is  obtained  by  putting  for  the  second  column  the  second 
members  of  the  given  equations. 


Therefore, 


3, 

28, 

7 

2, 

-23, 

-9 

4, 

9, 

-5 

3, 

-    &, 

7 

2, 

6, 

-9 

4, 

-    2, 

-5 

630 
:  -  210 : 


-3. 


EXERCISE   74 

Solve  by  determinants,  checking  each  result : 

3x  +  5y  =  l. 


2. 


5- 


i: 


2x  +  7y  =  -39. 
x  +  2y  =  ll. 


I5x  +  lly  =  82. 
.9x-Uy  =  8. 

r2x-y  +  3z  =  9. 
L  +  2y-z  =  2. 
^3x  +  5y  —  4z  =  L 

t5x  —  ly  +  z  =  —  1. 
2a  +  8?/-7z  =  -26. 

la  +  2*/  +  4z  =  4. 


8. 


9  a  +  5  z  =  —  7. 
9?/  +  3z  =  2. 
4#  —  11 2/  —  52  =  9. 
8a?  +  4y  —  2  =  11. 
16#  +  7?/-f-6z  =  64. 
2x  —  y  +  z  =  —  9. 
a  —  2?/  +  z  =  0. 
La-y  +  2«  =  -ll. 

ra  +  2?/-3z  =  5. 
3a-22?/  +  6z  =  4. 
,7a-6y-3«  =  15. 


6a;  —  4i/  —  7  2  =  17. 

[5  +  ^  =  2&. 
a     c 

9x-7y-16z  =  29. 
10x-5y-3z  =  23. 

10. 

?  +  *  =  2a. 
6      c 

^o      a 

230  ALGEBRA 

XIV.    THEORY   OF   EQUATIONS 

312.  Every  equation  of  the  nth  degree,  with  one  unknown 
number,  can  be  written  in  the  form 

xn  4-  PjZ"-1  +p»xn-2  +  •••  +pn-iX  +pn  =  0 ;  (1) 

where  the  coefficients  may  be  positive  or  negative,  integral  or 
fractional,  rational  or  irrational,  real  or  imaginary,  or  zero. 

If  no  coefficient  equals  zero,  the  equation  is  said  to  be  Com- 
plete; otherwise  it  is  said  to  be  Incomplete. 

We  shall  call  (1)  the  General  Form  of  the  equation  of  the  nth 
degree. 

313.  We  assume  that  every  equation  of  the  above  form  has 
at  least  one  root,  real  or  imaginary. 

314.  Divisibility  of  Equations. 

It  follows,  from  §  105,  that  if  the  equation 

Xn  +  p^-1  +    •  •  •  +  Pn-1%  +  Pn  =  0 

has  a  as  a  root,  then  the  first  member  is  divisible  by  x  —  a. 

For,  if  a  is  a  root,  the  first  member  becomes  0  when  x  is  put 
equal  to  a. 

315.  (Converse  of  §  314.)     If  the  first  member  of  the  equation 

xn  +  ppf*1  +  —  +  Pn-l®  +  Pn  =  0 

is  divisible  by  x  —  a,  then  a  is  a  root  of  the  equation. 

For  since  the  first  member  of  the  given  equation  is  divisible 
by  x  —  a,  the  equation  may  be  put  in  the  form 

(*-a)Q  =  0; 

and  it  follows  from  §  110  that  a  is  a  root  of  this  equation. 
It  follows  from  the  above  that  if  the  first  member  of 
IW?  +  Pi&~l+  •"  +Pn-lX+Pn  =  0 

is  divisible  by  ax+  6,  then is  a  root  of  the  equation. 

a 


THEORY  OF  EQUATIONS  231 

316.    Number  of  Roots. 

An  equation  of  the  nth  degree  has  n  roots,  and  not 
more  than  n. 

Let  the  equation  be 

Xn  +pYXn-^  +p2Xn~2  H hPn-lX  +Pn  =  0.  (1) 

By  §  313,  this  equation  has  at  least  one  root. 
Let  a  be  this  root;  then,  by  §  314,  the  first  member  is  divis- 
ible by  x  —  a,  and  the  equation  may  be  put  in  the  form 

(x-a)(xn^+q2xn-2+  ...  +  qn~lx+qn)  ==  0. 

By  §  110,  the  latter  equation  may  be  solved  by  placing 

x  —  a  =  0, 

and  x"-1 + q>xn~2  H \-q  n_x  x+qn  =  0.  (2) 

Equation  (2)  must  also  have  at  least  one  root. 
Let  b  be  this  root ;  then  (2)  may  be  written 

(x-b)  (xn~2  +  r3xn~s  +  •  •  •  +  *Li?  +  rn)  =  0, 

and  the  equation  may  be  solved  by  placing 

x  -  b  =  0, 

and  xn~2  -f  rsxn~3  -\ h  rn  xx  +  rn  =  0. 

After  Ti  —  1  binomial  factors  have  been  divided  out,  we  shall 
arrive  finally  at  an  equation  of  the  first  degree, 

x  —  7c  =  0 ;  whence,  x  =  7c. 

Therefore,  the  given  equation  has  the  n  roots  a,  6,  •  •  •,  7c. 

The  roots  are  not  necessarily  unequal. 

m 

Ex.  x3  -  3  x2  +  3  x  -  1  =  0. 

Whence,  (x  -  l)(x  -  l)(x  - 1)  =  0 

and  x  =  l,  or  1,  or  1. 


232  ALGEBRA 

317.    Depression  of  Equations. 

It  follows  from  §  316  that,  if  m  roots  of  an  equation  of  the 
nth  degree  are  known,  the  equation  may  be  depressed  to  an- 
other equation  of  the  (n  —  m)th  degree,  which  shall  contain  the 
other  n  —  m  roots. 

Thus,  if  all  but  two  of  the  roots  of  an  equation  are  known, 
these  two  may  be  obtained  from  the  depressed  equation  by  the 
rules  for  quadratics. 

Ex.  Two  roots  of  the  equation  9  x4  —  37  x2  —  8  x  +  20  =  0  are 
2  and  —  f  ;  find  the  others. 

By  §  314,  the  first  member  of  the  given  equation  is  divisible  by 
(a- 2)  (3«  +  5),  or  Sx2  -  x  -  10. 

Dividing  9  x4  —  37  x2  —  8  x  +  20  by  3  x2  —  x  —  10,  the  quotient  is 
3x2  +  x-2. 

Then  the  depressed  equation  is 

3  x2  +  x  -  2  =  0. 
Solving  by  the  rules  for  quadratics,  x  =  }  or  —  1. 

EXERCISE   75 

Find  whether : 

i.  One  root  of  ar3  —  x2  —  32  x  +  60  =  0  is  5;  if  so,  find  the 
others. 

2.  One  root  of  2  a?  -  6  x2  -  2  x  +  6  =  0  is  3 ;  if  so,  find  the 
others. 

3.  Two  roots  of  4  x4  -  12  x3  -  13  x2  +  45  x  - 18  =  0  are  -  2 

and  3  respectively ;  if  so,  find  the  others. 

4.  One  root  of  3  Xs  —  5  x2  +  2  x  —  4  =  0  is  2 ;  if  so,  find  the 
others. 

5.  Two  roots  of  14  a4  +  65  ar*  -  222  x2  +  65  x  + 14  =  0  are  2 
and  —  7  ;  if  so,  find  the  others. 

6.  Two  roots  of  x4  +  4  xs  -  6  x2  +  24  x  -  72  =  0  are  -  6  and 
2;    if  so,  find  the  others. 

7.  Two  roots  of  x4  —  4  Xs  +  3  x2  +  4  a;  —  4  =  0  are  2  and  —  1 ; 
if  so,  find  the  others. 


THEORY  OF  EQUATIONS  233 

8.  Three  roots  of  2  x*  +  29  x4  +  1*8  or3  +  379  a;2  +  394  x  + 120 
=  0  are  —  2,  —  3,  —  £  j  if  so,  find  the  others. 

9.  Two  roots  of  a*4  +  12  a8  +  34  x2  -  12  x  -  35  =  0  are  1  and 
—  7  ;  if  so,  find  the  others. 

10.    Two  roots  of  5  x*  -  18  a)3  +72z  -  120  =  0  are  5  and  -  4  ; 
if  so,  find  the  others. 

318.   Formation  of  Equations. 

It  follows  from  §  316,  that  if  the  roots  of 

xn+pxxn~l-\ t-Pn_lX+Pn=zO 

are  a,  b,  •••,  k,  the  equation  may  be  written  in  the  form 

(x  —  a)(x  —  b)  •  •  •  (x  —  k)  =  0. 

Hence,  to  form  an  equation  which  shall  have  any  required 
roots, 

Subtract  each  root  from  a?,  and  place  the  product  of 
the  resulting  expressions  equal  to  zero. 

Ex.   Form  an  equation  having  the  roots  1,  i,  and  —  f . 

By  the  rule  (x-  l)(x  -DO  +  })  =  0. 

Multiplying  the  terms  of  the  second  factor  by  2,  and  of  the  third  by  3, 

(x-l)(2z-l)(3£  +  5)  =0. 
Expanding,  6x3  +  %2  —  12  x  +  5  =  0. 

EXERCISE   76 

Form  equations  having  the  roots  : 

I-  1,-4,6.  7.  _m)m±v^. 

2.   3,-1,-21  4 

3-2,3,5,0.  8.  3±V2,-3±V2. 

4.    -1,-2,7,-8.  9-   «,  -~>-&>-J- 

5-   2,  9,  -  J,  |.  ^    4±2V3    -2±V3 

6.   4,4,  -I,  -£.  3       '  3 


234  ALGEBRA 

319.    Composition  of  Coefficients. 

By  §  318,  the  equation  of  the  nth  degree  whose  roots  are  a, 
b,  c,  d,  •  ••,  ft,  Z,  m,  is 

(x  —  a)(x  —  b)(x  —  c)(x  —  d)  •••  (x  —  m)  =  0.  (1) 

By  actual  multiplication,  we  obtain 

(x  —  a)  (x  —  b)  —  x2  —  (a  +  b)x  +  ab  ; 
(x  —  a)  (x  —  b)  (x  —  c) 

=  x*  —  («  +  &  +  c)x2  +  (ab  -f  be  -f  ca)x  —  a&c ; 
(#  —  a) (x  —  6)  (x  —  c)  (x  —  d) 

=  xi—(a  +  b  +  c  +  d)x*  -f  (a&  +  ac+  ad +  bc  +  bd  +  cd)x2 

—  (abc  +  abd  +  aceZ  +  6cd)x  +  abed  =  0  ;  etc. 

When  all  the  factors  of  the  first  member  of  (1)  have  been  multiplied 
together,  the  result  will  be  in  the  form 

xn  -\-piXn~l  +  p2xn-2  +  p3xn-s  -f  •••  +pn; 

where  pi  =  -  (a  +  b  -f  c  +  •••  +  k  +  I  -f  m)  ; 

p2  =  ab  +  ac  +  6c  +  •  •  •  -f  Zm  ; 

p3  =  —  (abc  +  «6c?  +  acd  +  •••  +  Mm); 


pn  =  ±  abed  •••  klm,  according  as  n  is  even  or  odd. 
Hence,  in  an  equation  of  the  nth  degree  in  the  general  form, 

The  coefficient  of  the  second  term  is  equal  to  minus 
the  sum  of  all  the  roots. 

The  coefficient  of  the  third  term  is  equal  to  the  sum  of 
the  products  of  the  roots,  taken  two  at  a  time, 

The  coefficient  of  the  fourth  term  is  equal  to  minus  the 
sum  of  the  products  of  the  roots,  taken  three  at  a  time ; 
etc. 

The  last  term  is  equal  to  plus  or  minus  the  product  of 
all  the  roots,  according  as  n  is  even  or  odd. 

320.  It  follows  from  §  319  that,  if  an  equation  of  the  nth 
degree  is  in  the  general  form, 

If  the  second  term  is  wanting,  the  sum  of  the  roots  is  0. 
If  the  last  term  is  wanting,  at  least  one  root  is  0. 


THEORY   OF   EQUATIONS  235 

If  the  last  term  is  an  integer,  it  is  divisible  by  every  integral 
root. 

EXERCISE  77 

In  each  of  the  following,  find  the  sum  of  the  roots,  and  the 
product  of  the  roots : 

i.   xz-Sx2  +  l§x-12  =  b.  2.   ar3-  31  x-  30  =  0. 

3.   4^-12^  +  3^  +  13^-6=0. 

321.  If  all  but  one  of  the  roots  of  an  equation  of  the  ?ith 
degree  in  the  general  form  are  known,  the  remaining  root  may 
be  found  by  changing  the  sign  of  the  coefficient  of  the  second 
term  of  the  given  equation,  and  subtracting  the  sum  of  the 
known  roots  from  the  result ;  or,  by  dividing  the  last  term  of 
the  given  equation  if  n  is  even,  or  its  negative  if  n  is  odd,  by 
the  product  of  the  known  roots. 

If  all  but  two  are  known,  the  coefficient  of  the  second  term 
of  the  depressed  equation  may  be  found  by  adding  the  sum  of 
the  known  roots  to  the  coefficient  of  the  second  term  of  the 
given  equation ;  and  the  last  term  of  the  depressed  equation 
may  be  found  by  dividing  the  last  term  of  the  given  equation 
by  plus  or  minus  the  product  of  the  known  roots  according  as 
n  is  even  or  odd. 

Ex.  Two  roots  of  the  equation  9  xA  —  37  x2  -  8  a; +  20  =  0 
are  2  and  —  f ;  what  are  the  others  ? 

We  first  put  the  equation  in  the  general  form  by  dividing  each  term 
by  9. 

It  then  becomes  sc*  -  -3/  x2  —  §  x  +  ^  =  0. 

Since  there  is  no  sc3  term,  the  coefficient  of  the  second  term  is  0. 

Then  the  coefficient  of  the  second  term  of  the  depressed  equation  is 
0  +  2  -  §  or  1 

The  coefficient  of  the  last  term  of  the  depressed  equation  is 


Solving,  x  =  §  or  —  1. 


236  ALGEBRA 

EXERCISE   78 

i.   One  root  of  x3  +  7  x2  —  5x  —  75  =  0  is  —  5 ;  find  the  others. 

2.  Three  roots  of  4  a;4  —  55  x2  —  45  #  +  36  =  0  are  4,  —3,  \\ 
find  the  other. 

3.  Four  roots  of  20  x5  -  108  x4  +  225  or3  -  224  x2  + 105  a  -  18 
=  0  are  1,  1,  -§,  f ;  find  the  other. 

4.  Three  roots  of  or5  -  a4  -  15  Xs  + 25  ar>  + 14  x  -  24  =  0  are  2, 
1,  —  4 ;  find  the  others. 

5.  Two  roots  of  x*  -  ax*  +  (2  a  -  7  a2  -  l)rf  +  (a3  -  2  a2  +  a)* 
+  6  a4  — 12  a3  +  6  a2  =  0  are  a  —  1  and  3  a;  find  the  others. 

322.    Fractional  Roots. 

An  equation  in  the  general  form  with  integral  coeffi- 
cients cannot  have  as  a  root  a  rational  fraction  in  its 
lowest  terms. 

Let  the  equation  be 

xn  +p1aft~1  +p2xn  2  H VPn-l*  +Pn  =  0, 

where  pl9  p2,  '"}pn  are  integral. 

If  possible,  let  -,  a  rational  fraction  in  its  lowest  terms,  be  a  root  of 
b 

the  equation ;  then, 

Multiplying  each  term  by  6n_1,  and  transposing, 

b 

By  hypothesis,  a  and  b  have  no  common  divisor  ;  hence,  an  and  b  have 
no  common  divisor. 

We  then  have  a  rational  fraction  in  its  lowest  terms  equal  to  an  integral 
expression,  which  is  impossible. 

Therefore,  the  equation  cannot  have  as  a  root  a  rational  fraction  in  its 
lowest  terms. 


THEORY   OF   EQUATIONS  237 

323.    Imaginary  Roots. 

If  the  imaginary  number  a  +  hi  is  a  root  of  an  equation 
in  the  general  form,  with  real  coefficients,  its  conjugate 
(a  -  bi)  is  also  a  root. 

Let  the  equation  be 

Xn  +plXn-l  +   . . .  +pn_lX  +pn  =  0,  (1) 

where  p1?  •••,  pn  are  real  numbers. 

Since  a  +  bi  is  a  root  of  (1),  we  have 

(a  +  bi)»  +  pi(a  +  bi)"-1  +  •-  +Pn-i{a+bi)  +  pn  =  0. 

Expanding  by  the  Binomial  Theorem,  we  have,  by  §  180, 

an  +  nan-m  _  riin-J^       %2_  n(n  -  l)(n  -  2)  _ 

[2  >(S 

+  pjan~l  +  (n-  \)a-m  -  ^  -  *) 0*  -  2)  aW_3&2 1 

+  ...  +j£»-i(a  +  50  +pw  =  0.  (2) 

Collecting  the  real  and  imaginarj^  terms,  we  have  a  result  of  the  form 

P+Qi  =  0.  (3) 

Here,  P  stands  for  the  sum  of  all  the  terms  containing  a  alone,  together 
with  all  the  terms  containing  even  powers  of  i ;  and  Qi  for  all  terms 
containing  odd  powers  of  i. 

In  order  that  equation  (3)  may  hold,  we  must  have 

P  =  0,  and  Q  =  0. 

Now  substituting  a  —  bi  for  x  in  the  first  member  of  equation  (1), 
it  becomes 

(a  -  60"  +Pi(a  -  bi)*-1  +  -  +pn-i(a  -  bi)  fpn.  (4) 

Expanding  by  the  Binomial  Theorem,  we  shall  have  a  result  which 
differs  from  the  first  member  of  (2)  only  in  having  the  second,  fourth, 
sixth,  etc.,  terms  of  each  expansion,  or  those  involving  i  as  a  factor, 
changed  in  sign. 

Then,  collecting  the  real  and  imaginary  terms,  the  expression  (3)  is 
equal  to  p  _  q^ 

where  P  and  Q  have  the  same  meanings  as  before. 
But  since  P  =  0  and  Q  *=  0,  we  have  P—  Qi  =s  0. 
Whence,  a  —  bi  is  a  root  of  (1).  *• 

The  above  proof  holds  without  change  when  a  equals  zero  ;  thus  the 

theorem  holds  for  any  pure  imaginary  number,  of  the  form  bi. 


238  ALGEBRA 

324.  The  product  of  the  factors  of  the  first  member  of 
equation  (1),  §  323,  corresponding  to  the  conjugate  imaginary 
roots  a+bi  and  a  —  bi  is 

[x  _  (a  +  bi)J_x -  (a-bi)-]  (§  318) 

=  (x  —  a  —  bi)(x  —  a  +  bi) 

=  (x-  a)2  -  (bi)2  =  (x  -  a)2  +  b2 ; 

and  is  therefore  positive  for  every  real  value  of  x. 

325.  It  follows  from  §§  316  and  323  that  every  equation  of 
odd  degree  has  at  least  one  real  root ;  for  an  equation  cannot 
have  an  odd  number  of  imaginary  roots. 

TRANSFORMATION   OF   EQUATIONS 

326.  To  transform  an  equation  into  another  which  shall  have 
the  same  roots  with  contrary  signs. 

Let  the  equation  be 

xn  +Pifl  +p2xn~2  +  •••  +  Pn-iX  +pn  =  0.  (1) 

Substituting  —  y  for  as,  we  have 

(-  y)n  +pi(-  y)71-1  +m-  y)n~2  +  -  +p»-i(- y)  +pn  =  o. 

Dividing  each  term  by  (—  1)M,  we  have 

Or,  yn_piyn-l  +  p2yn-2 ±  Pn-l2/  T  Pn  =  0  J  (2) 

the  upper  or  lower  signs  being  taken  according  as  n  is  odd  or  even. 

It  follows  from  (1)  and  (2)  that  the  desired  transformation 
may  be  effected  by  simply  changing  the  signs  of  the  alternate 
terms  commencing  with  the  second. 

If  the  equation  is  incomplete.,  any  missing  term  must  be  supplied  with 
the  coefficient  zero  before  applying  the  rule. 


THEORY  OF  EQUATIONS  239 

327.  Ex.  Transform  the  equation  a;3  —  10  x  +  4  =  0  into 
another  which  shall  have  the  same  roots  with  contrary 
signs. 

The  equation  may  be  written  xs  +  0  •  x2  —  10  x  +  4  =  0. 
Then,  by  the  rule,  the  transformed  equation  is 

X3_0.x2-  10x-4  =  0,  or  x3-10x-4=0. 

EXERCISE   79 

Transform  each  of  the  following  into  an  equation  which 
shall  have  the  same  roots  with  contrary  signs : 

i.   xs-.6x2  +  12x-8  =  0.     2.   x*-6x3  +  ±x2-9x  +  16  =  0. 

3.   x7  +  5x5  —  3xi  +  x  —  4:  =  0. 

328.  To  transform  an  equation  into  another  whose  roots  shall 
be  respectively  m  times  those  of  the  first. 

Let  the  equation  be 

Xn  +^>1Xn~1  -\-p2Xn~2  +   •  •  •  +Pn-iX  +pn  =  0. 

Putting  mx  =  y,  that  is,  2-  for  x,  we  have 


m 


+  Pn  =  0. 


Multiplying  each  term  by  mn, 

yn  -\-piimyn-1  -\-p2m2yn-2  +  ...  + pn-\mn-ly  +  pnmn  =  0. 

Hence,   to   effect  the  desired  transformation,  multiply  the 
second  term  by  m,  the  third  term  by  m2,  and  so  on. 

'    Ex.     Transform  the  equation  Xs  -\-7  x2  —  6  =  0  into  another 
whose  roots  shall  be  respectively  4  times  those  of  the  first. 

Supplying  the  missing  term  with  the  coefficient  zero,  and  applying  the 
rule,  we  have 

x3  +  4  .  7  x1  +  4*  •  Ox  -  4«  •  6  =  0,  or  x3  +  28  x2  -  384  =  0. 


240  ALGEBRA 

329.  To  transform  an  equation  with  fractional  coefficients  into 
another  whose  coefficients  shall  be  integral,  that  of  the  first  term 
being  unity. 

The  transformation  may  be  effected  by  transforming  the 
equation  into  another  whose  roots  shall  be  respectively  m 
times  those  of  the  first  (§  328) ;  we  then  give  to  m  such  a 
value  as  will  make  every  coefficient  integral. 

By  giving  to  m  the  least  value  which  will  make  every  coeffi- 
cient integral,  the  result  will  be  obtained  in  its  simplest  form. 

Ex.     Transform  the  equation  Xs — =  0  into  an- 

H  3      36     108 

other  whose  coefficients  shall  be  integral,  that  of  the -first  term 

being  unity. 

By  §  328,  the  equation 

^."V-^  +  ^O 
3  36         108 

has  its  roots  respectively  m  times  those  of  the  given  equation. 

It  is  evident,  by  inspection,  that  the  least  value  of  m  which  will  make 
every  coefficient  integral,  is  6. 

Putting  m  =  6,  we  have 

xs  _  2  x2  -  x  +  2  =  0, 
whose  roots  are  6  times  those  of  the  given  equation. 

330.  To  transform  an  equation  into  another  whose  roots  shall 
be  respectively  those  of  the  first  increased  by  m. 

Let  the  equation  be 

xn  +Plxn~l  +  .  •  •  +pn_xx  +pH  =  0.  (1) 

Putting  x  -f  m  =  y,  that  is,  y  —  m  for  x,  we  have 

(y  —  m)n+pi(y  —  m)n-l+ \-Pn-i(y  —  m)+pn  =  0.  (2) 

Expanding  the  powers  of  y  —  m  by  the  Binomial  Theorem,  and  collect- 
ing the  terms  involving  like  powers  of  y,  we  shall  have  a  result  of  the 
yn  +  qiyn-i  _|_  .. .  _+_  qn_iy  +  Qn  =  o,  (3) 

whose  roots  are  respectively  those  of  the  given  equation  increased  by  m. 


THEORY   OF    EQUATIONS  241 

Ex.  Transform  the  equation  or3  —  7  #  -f  6  =  0  into  another 
whose  roots  shall  be  respectively  those  of  the  first  increased  by  2. 

Substituting  y  —  2  for  cc, 

(j/-2)3-7G,-2)  +  6  =  0. 
Expanding,  and  collecting  the  terms  involving  like  powers  of  y,  we 
have  y8-6y*  +  6y  +  12  =  0. 

331.  If  m  and  the  coefficients  of  the  given  equation  are 
integral,  the  coefficients  of  the  transformed  equation  may  be 
conveniently  found  by  the  following  method. 

Putting  x  +  m  for  y  in  (3),  we  obtain 

(x  +  m)n  +  qi  (x  +  m)»  *  +  •••  +  g„_i(x  +  m)  +  qn  =  0,  (4) 

which  must,  of  course,  take   the  same  form  as  (1)  on   expanding  the 
powers  of  x  +  m,  and  collecting  the  terms  involving  like  powers  of  x. 
Dividing  the  first  member  of  (4)  by  x  +  w,  we  have 

(x  +  m)"-1  +  qx(x  +  m)»-a  -f  .-.  +q*-%  C*  +  w)  +  g„_i  (5) 

as  a  quotient,  with  a  remainder  qn. 

Dividing  (5)  by  x  +  m,  we  have  the  remainder  qn-i;  etc. 

Hence,  to  obtain  the  coefficients  of  the  transformed  equation : 

Divide  the  first  member  of  the  given  equation  by 
oo  +  m  ;  the  remainder  will  be  the  last  term  of  the  required 
equation. 

Divide  the  quotient  just  found  by  oc+  in  ;  the  remainder 
will  be  the  coefficient  of  the  next  to  the  last  term  of  the 
transformed  equation ;  and  so  on. 

Ex.  Transform  the  equation  sc3  — 7#4-6  =  0  into  another 
whose  roots  shall  be  respectively  those  of  the  first  increased  by  2. 

Dividing  xs  —  7  x  +  6  by  x  +  2,  we  have  the  quotient  x2  —  2  x  —  3,  and 
the  remainder  12  (§  108). 

Dividing  x2  —  2  x  —  3  by  x  +  2,  we  have  the  quotient  x  —  4,  and  the 
remainder  5.  N 

Dividing  x  —  4  by  x  +  2,  we  have  the  remainder  —  6. 

Then,  the  transformed  equation  is 

X3_6x2  +  5x  +  12  =  0. 
Compare  Ex.,  §  330. 


242  ALGEBRA 

332.  To  transform  an  equation  into  another  whose  roots 
shall  be  those  of  the  first  diminished  by  ra,  we  change  y  —  m  to 
y  -\-m  in  the  method  of  §  330,  and  x  -f-  m  to  x  —  m  in  the  rule 
of  §  331. 

EXERCISE  80 

i.  Transform  x2  —  x  —  12  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  5  times  the  first.     Verify  your  results. 

2.  Transform  ar3  -f-  x2  —  14  x  —  24  =  0  into  an  equation  whose 
roots  shall  be,  respectively,  twice  those  of  the  first.  Verify 
results. 

3.  Transform  x^-^Sx2  —  23  x  —  210  =  0  into  an  equation 
whose  roots  shall  be,  respectively,  \  times  the  first. 

Transform  each  of  the  following  into  an  equation  with  in- 
tegral coefficients,  that  of  the  first  term  being  unity  : 

4.  6  or3  -11  x2  -14  a; +  24  =  0.     Verify  result. 

5.  8x*  +  Ux2-5x-2  =  Q. 

6.  2a4-13ar3-91arJ  +  390a  +  216  =  0. 

7.  90^-flll^  +  25^2-12cc-4  =  0. 

8.  xA  +  —  -  —  -—  =  0. 

7        14      196 

9.  Transform  a^-f-lO  x2  +  7  x— 18  =  0  into  an  equation  whose 
roots  shall  be,  respectively,  those  of  the  first  diminished  by  4. 

10.  Transform  x*  -  3  Xs  — 19  x2  +  27  x  +  90  =  0  into  an  equa- 
tion whose  roots  shall  be,  respectively,  those  of  the  first  in- 
creased by  3. 

333.  To  transform  the  equation 

xn  +p1xn~1  -\ f-  pn_iX  -f  pn  =  0 

where  pl   is  not  zero,  into  another  whose  second  term   nkatt  be 
wanting. 


THEORY   OF   EQUATIONS  243 

Expanding  the  powers  of  y  —  m  in  the  first  member  of  (2), 
§  330,  and  collecting  the  terms  involving  like  powers  of  y,  we 
have 

yn  +  (l\  —  mn)yn~l-\ =  0. 

If  m  be  so  taken  that  pY  —  mn  =  0,  whence  m  =  — ,  the  coeffi- 

n 
cient  of  yn~^  will  be  zero. 

Hence,  the  desired  transformation  may  be  effected  by  sub- 
stituting in  the  given  equation  y  —  -in  place  of  x. 

Ex.  Transform  x5  —  6x2  +  9  x  —  6  =  0  into  an  equation  whose 
second  term  shall  be  wanting. 

_  ft 

Substituting  y or  y  +  2,  in  place  of  x,  we  have 

o 

(?/  +  2)»  _  6  (y  +  2)2  +  9  (y  +  2)  -  6  =  0. 
Then,  ys  +  6  y*  +  12  y  +  8  -  6  ?/2  -  24  y  -  24  +  9  y  +  18  -  6  =  0, 
or  y3  —  3  y  —  4  =  0  ; 

whose  roots  are  those  of  the  given  equation  diminished  by  2. 

EXERCISE  81 

Transform  each  of  the  following  into,  an  equation  whose 
second  term  shall  be  wanting : 

i.   ^-6a2  +  4a-l=0.  3.    x4  + 12 ^+2^-3  =  0. 

2.   x3  +  5x2  +  8  =  0.  4.    x5-xi  +  7x-l  =  0. 

DESCARTES'   RULE  OF   SIGNS 

334.  If  an  equation  of  the  nth  degree  is  in  the  general  form 
(§  312),  a  Permanence  of  sign  occurs  when  two  successive  terms 
have  the  same  sign,  and  a  Variation  of  sign  occurs  when  two 
successive  terms  have  opposite  signs. 

Thus,  in  the  equation  xG  —  3  x4  —  Xs  -f  5  x  + 1  =  0,  there  are 
0  permanences  and  two  variations. 


, 


244  ALGEBRA 

335.    Descartes'  Rule  of  Signs. 

No  equation,  whether  complete  or  incomplete,  can 
have  a  greater  number  of  positive  roots  than  it  has 
variations  of  sign ;  and  no  complete  equation  can  have 
a  greater  number  of  negative  roots  than  it  has  perma- 
nences of  sign. 

Let  an  equation  in  the  general  form  have  the  following  signs : 

+  +  0  -  +  00 , 

the  missing  terms  being  supplied  with  zero  coefficients. 

If  we  introduce  a  new  positive  root  a,  we  multiply  this  by  x  —  a  (§  318) ; 
writing  only  the  signs  which  occur  in  the  process,  we  have 

123456789 

+     +0-+00--  (1) 

+    - 


+ 

+ 

0 

— 

+ 

0    0 

— 

— 

— 

— 

0 

+ 

-    0 

0 

+ 

+ 

+ 

m 

— 

— 

+ 

-    0 

— 

m 

+ 

1 

2 

3 

4 

5 

6     7 

8 

9 

10 

(2) 

9      10 

Here  m  signifies  a  term  which  may  be  +,  0,  or  — . 

Now,  in  (1),  let  a  dot  be  placed  over  the  first  minus  sign,  then  over 
the  next  plus  sign,  then  over  the  next  minus  sign,  and  so  on. 

The  number  of  dots  shows  the  number  of  variations  ;  thus  in  (1)  there 
are  three  variations. 

In  the  above  result,  we  observe  the  following  laws : 

I.  Directly  under  each  dotted  term  of  (1)  is  a  term  of  (2) 
having  the  same  sign. 

Thus,  the  terms  numbered  4,  5,  and  8,  in  (1)  and  (2),  have 
the  same  sign. 

II.  The  last  term  of  (2)  is  of  opposite  sign  to  the  term 
directly  under  the  last  dotted  term  of  (1). 

The  above  laws  are  easily  seen  to  hold  universally. 

By  the  first  law,  however  the  term  marked  m  is  taken,  there 
are  at  least  as  many  variations  in  the  first  eight  terms  of  (2) 
as  in  (1)  ;  and  by  the  second  law,  there  is  at  least  one  varia- 
tion in  the  remaining  terms  of  (2). 

Hence,  the  introduction  of  a  new  positive  root  increases  the 
number  of  variations  in  the  equation  by  at  least  one. 


THEORY  OF  EQUATIONS  245 

If,  then,  we  form  the  product  of  all  the  factors  correspond- 
ing to  the  negative  and  imaginary  roots  of  an  equation,  multi- 
plying the  result  by  the  factor  corresponding  to  each  positive 
root  introduces  at  least  one  variation. 

Hence,  the  equation  cannot  have  a  greater  number  of  posi- 
tive roots  than  it  has  variations  of  sign. 

To  prove  the  second  part  of  Descartes'  Rule,  let  —  y  be  sub- 
stituted for  x  in  any  complete  equation. 

Then  since  the  signs  of  the  alternate  terms  commencing 
with  the  second  are  changed  (§  326),  the  original  permanences 
of  sign  become  variations. 

But  the  transformed  equation  cannot  have  a  greater  number 
of  positive  roots  than  it  has  variations. 

Hence,  the  original  equation  cannot  have  a  greater  number 
of  negative  roots  than  it  has  permanences. 

In  all  applications  of  Descartes'  Rule,  the  equation  must  contain  a 
term  independent  of  x  ;  that  is,  no  root  must  equal  zero  ;  for  a  zero  root 
cannot  be  regarded  as  either  positive  or  negative. 

336.  It  follows  from  the  last  part  of  §  335,  and  from  §  326, 
that  in  any  equation,  whether  complete  or  incomplete,  the 
number  of  negative  roots  cannot  exceed  the  number  of  varia- 
tions in  the  equation  which  is  formed  from  the  given  equation 
by  changing  the  signs  of  the  alternate  terms  commencing  with 
the  second. 

337.  In  any  complete  equation,  the  sum  of  the  number  of 
permanences  and  variations  is  equal  to  the  number  of  terms 
less  one,  or  to  the  degree  of  the  equation. 

That  is,  the  sum  of  the  number  of  permanences  and  varia- 
tions is  equal  to  the  number  of  roots  (§  316). 

Hence,  if  the  roots  of  a  complete  equation  are  all  real,  the 
number  of  positive  roots  equals  the  number  of  variations,  and  the 
number  of  negative  roots  equals  the  number  of  permanences. 

An  equation  whose  terms  are  all  positive  can  have  »o  posi- 
tive root ;  and  a  complete  equation  whose  terms  are  alternately 
positive  and  negative  can  have  no  negative  root. 


246  ALGEBRA 

338.   Ex.   Determine  the  nature  of  the  roots  of 
x3  +  2  x  +  5  =  0. 

There  is  no  variation,  and  consequently  no  positive  root. 

Changing  the  signs  of  the  alternate  terms  commencing  with  the  second, 
we  have  xs  +  2  x  -  5  =  0. .    (See  Note,  §  326. ) 

Here  there  is  one  variation  ;  and  therefore  the  given  equation  cannot 
have  more  than  one  negative  root  (§  330). 

Then  since  the  equation  has  three  roots  (§  316),  one  of  them  must  be 
negative  and  the  other  two  imaginary. 

If  two  or  more  successive  terms  of  an  equation  are  wanting,  it  follows 
by  Descartes'  Rule  that  the  equation  must  have  imaginary  roots. 

EXERCISE   82 

If  the  roots  of  the  following  are  all  real,  determine  their 
SlgnS:  i.  ^  +  10^  +  7^-18  =  0. 

2.  a4-3ar3-19#2  +  27a  +  90=0. 

3.  36x?-mxs  +  '27x2  +  7x-3  =  0. 

4.  ar5-4a4-5a3  +  20a;2  +  4a-16=0. 

5.  2a4-13arJ-91a2  +  390a  +  2iG  =  0. 

Determine  the  nature  of  the  roots  of  the  following : 

6.  2^  +  0^+2^-12  =  0. 

7.  #4  +  3ar?  +  7arJ  +  6a;  +  4  =  0. 

8.  a4-2^-9  =  0. 

9.  x>-2x4  +  4xs-8x2  +  16x-16=0. 
io.  x7  +  3  x4  +  5x2  +  2  =  0. 

LIMITS   TO   THE   ROOTS 

339.    To  find  a  superior  limit  to   the  positive   roots  oj   an 
equation. 

The  following  examples  illustrate  the  method  of  finding  a 
superior  limit  to  the  positive  roots  of  an  equation. 


• 


THEORY   OF    EQUATIONS  247 

1.  Find  a  superior  limit  to  the  positive  roots  of 

x*  -  3  x2  +  2  x  -  5  =  0. 

Grouping  the  positive  and  negative  terms,  we  can  write  the  first  mem- 
ber in  the  form 

a*(a-3)+2(a:- J).  (1) 

It  is  evident  that  if  x  equals  or  exceeds  8,  the  expression  (1)  is  positive 
Hence,  no  root  of  the  given  equation  equals  or  exceeds  3,  and  8  is  a 
superior  limit  to  the  positive  roots. 

2.  Find  a  superior  limit  to  the  positive  roots  of 

a4 -15  a2- 10  a+ 24  =  0. 

2  x^         x^ 
We  separate  the  first  term  into  the  parts  -r-  and  — ,  and  write  the  first 

member  in  the  form 

/?*!_L5  xA  +^_M)aA  +  24,  or  ^(2  x2  -  45)  +  -(a*  -  30)  +  24. 

It  is  evident  from  this  that  no  root  can  be  so  great  as  5 ;  hence,  5  is  a 
superior  limit  to  the  positive  roots. 

If  we  had  written  the  first  member  in  the  form 

^_  15^2 \  +  ht_  iox\  +24,  or  £-2(z2  -  30)  +  -<>3  -  20)  +  24,' 

we  should  have  found  6  as  a  superior  limit  to  the  positive  roots. 

2  x4         sc4  sc4         sc4 

Thus,  separating  x4  into  —  and  — ,  instead  of  —  and  — ,  gives  a  smaller 
So  A  A 

limit. 


340.  To  find  an  inferior  limit  to  the  negative  roots  of  an 
equation. 

First  transform  the  equation  into  another  which  shall  have 
the  same  roots  with  contrary  signs  (§  326). 

The  superior  limit  to  the  positive  roots  of  the -transformed 
equation,  obtained  as  in  §  339,  with  its  sign  changed,  will 
be  an  inferior  limit  to  the  negative  roots  of  the  given  equa- 
ion. 


248  ALGEBRA 

Ex.   Find  an  inferior  limit  to  the  negative  roots  of 

tf  +  2  x*  +  5  x2  -  7  =  0. 

Transforming  the  equation  into  another  which  shall  have  the  same 
roots  with  contrary  signs  (§  326),  we  have 

xb  +2z3-5z2-f  7  =  0.  (lj 

We  can  write  the  first  member  in  the  form 

x2(x3-5)  +  2z3  +  7. 

It  is  evident  from  this  that  no  root  of  (1)  can  be  so  great  as  2  ;  hence, 
—  2  is  an  inferior  limit  to  the  negative  roots  of  the  given  equation. 

By  grouping  the  xb  and  x2  terms  in  (1),  we  obtain  a  smaller  limit  than 
if  we  group  the  x3  and  x2  terms. 

EXERCISE  83 

In  each  of  the  following,  find  a  superior  limit  to  the  positive 
roots,  and  an  inferior  limit  to  the  negative : 

,  i.  arJ  +  3a2  +  a~-4  =  0. 

2.  xA  +  5x*-15x-9  =  0. 

3.  x4  +  3x*  -5  x  -8  =  0. 

4.  3x*-5x2-8x-7  =  0. 

5.  ^-4a4  +  6ar3  +  32a2__  15^  +  3  =  0. 

6.  2x5  +  5xi  +  6xi-13x2-25x  +  4;  =  0. 

7.  In  the  equation  Xs  —  2  x2  —  3a?  +  l  =  0,  prove  3  a  supe- 
rior limit  to  the  positive  roots,  and  —  2  an  inferior  limit  to  the 
negative. 

8.  In  the  equation  2  a8  +  5  aj2  —  7  x  —  3  =  0,  prove  —  4  an 
inferior  limit  to  the  negative  roots,  and  find  a  superior  limit  to 
the  positive. 

9.  In  the  equation  x4  -f  3  x*  —  9  x2  +  12  x  —  10  =  0,  prove  3 
a  superior  limit  to  the  positive  roots,  and  —  G  an  inferior  limit 
to  the  negative. 


THEORY  OF   EQUATIONS  249 

DERIVATIVES 

341.  If  we  take  the  polynomial 

axn  +  bx"-1  +  cxn~2  +  •••, 

multiply  each  term  by  the  exponent  of  x  in  that  term,  and 
then  diminish  the  exponent  by  1,  the  result 

naxn~l  -f  (n  —  l)bxn~2  +  (n  —  2)cxn~z  +  ... 

is  called  the  first  derivative  of  the  given  polynomial. 

The  first  derivative  of  the  first  derivative  is  called  the  sec- 
ond derivative  of  the  given  polynomial ;  the  first  derivative 
of  the  second  derivative  is  called  the  third  derivative  ;  and 
so  on. 

Ex.     Find  the  successive  derivatives  of  3  Xs— 9  x2  — 12  x  -f  2. 

The  first  is  9  x2  -  18  x  -  12. 
The  second  is  18  x—  18. 
The  third  is  18. 
The  fourth  is  0. 

It  will  be  understood  hereafter  that  when  we  speak  of  the  derivative  of 
an  expression,  we  mean  the  first  derivative. 

EXERCISE  84 

Find  the  successive  derivatives  of : 

i.   5x2  +  $x-7.  4    8^  -3  a2  +  2. 

2.  3^-7^  +  2.  5.   6x6-5x5  +  4:Xs-3x2  +  27. 

3.  9z3-7a2  +  15a;-l.  6.   x5  -  a?  4- 10  Xs  +  5  x2  -  7  x. 

MULTIPLE  ROOTS 

342.  If  an  equation  has  two  or  more  roots  equal  to  a,  a  is 
said  to  be  a  Multiple  Root  of  the  equation. 

In  the  above  case,  a  is  called  a  double  root,  a  tripte  root,  a 
quadruple  root,  etc.,  according  as  the  equation  has  two  roots, 
three  roots,  four  roots,  etc.,  equal  to  a. 


250  ALGEBRA 

343.  Let  the  roots  of  the  equation 

&+P&-* +&#*>+  -.  +pn  =  0  (1) 

be  a,  b,  c,  d,  •••. 

Then,  by  §>  318,  we  have 

xn  +  piic"-1  +  P2Xn~2  +  •••  =  (£-«)(£-  &)(£  — c)  •••. 

Putting  x  +  h  in  place  of  x,  we  obtain 

(x  +  ft)"  +pi(x  +  ft)*"1  +p2(x  +  ft)n~2  +  ••• 

=  (ft  +  x  -  a)  (ft  +  x  -  b)(h  +  x  -  c)  •••.  (2) 

Expanding  the  powers  of  x  +  ft  by  the  Binomial  Theorem,  the  coeffi- 
cient of  h  in  the  first  member  of  (2)  is 

nxn~l  +pi(n  —  l)xn~2  +p2(n  -  2)xn~s  +  •••  ;  (3) 

which,  we  observe,  is  the  first  derivative  of  the  first  member  of  (1). 

Again,  it  is  evident  from  §  319  that  the  coefficient  of  h  in  the  second 
member  of  (2)  is 

(x  —  b) (x  —  c) (x  —  d)  •••  to  n  —  1  factors 
+  (x  —  a)(x  —  c)(x  —  d)  •••  to  n  —  1  factors 
+  (x  —  a)  (x  —  b)  {x  —  d)  •  •  •  to  n  —  1  factors  +  •  •• .  (4) 

Since  equation  (2)  is  true  for  every  value  of  ft,  by  §  264  these  coeffi- 
cients of  ft  in  the  two  members  are  equal. 

Now  if  b  =  a,  that  is,  if  equation  (1)  has  two  roots  equal  to  «,  every 
term  of  (4)  will  be  divisible  by  x  —  a,  and  therefore  the  expression  (3) 
will  be  divisible  by  x  —  a. 

Hence,  the  equation  formed  by  equating  (3)  to  zero  will  have  one  root 
equal  to  a  (§  315). 

In  like  manner,  if  c  =  b  =  a,  that  is,  if  (1)  has  three  roots 
equal  to  a,  the  equation  formed  by  equating  (3)  to  zero  will 
have  two  roots  equal  to  a ;  and  so  on. 

Hence,  if  any  equation  of  the  form  (1)  has  m  roots 
equal  to  a,  the  equation  formed  by  equating  to  zero  the 
derivative  of  its  first  member  -will  have  m  —  1  roots  equal 
to  a. 

344.  It  follows  from  §  343  that,  to  determine  the  existence 
of  multiple  roots  in  an  equation  of  the  form 

PoXn+PlXn-1+    ••'    +Pn-\X+Pn  =  0, 

we  proceed  as  follows : 


THEORY  OF  EQUATIONS  251 

Find  the  II.  C.  F.  of  the  first  member  and  its  derivative. 
If  there  is  no  H.  C.  F.,  there  can  be  no  multiple  roots. 
If  there  is  a  II  C.  F.,  by  equating  it  to  zero  and  solving  the 
residting  equation,  the  required  roots  may  be  obtained. 

It  is  to  be  observed  that  the  number  of  times  that  each  root 
occurs  in  the  given  equation  exceeds  by  one  the  number  of 
times  that  it  occurs  in  the  equation  formed  by  equating  the 
H.  C.  F.  to  zero. 

Ex.    Find  all  the  roots  of 

a5 +0*  -  9 a? -  5a?  +  16s +  12  =  0.  (1) 

The  derivative  of  the  first  member  is 

5z4  +  4x3-27x2-10x4  16. 

The  H.  C.  F.  of  this  and  the  first  member  of  (1)  is  x2  —  x  —  2. 
Solving  the  equation  x2  —  x  —  2  =  0,  x  =  2  or  — 1. 
Then,  the  multiple  roots  of  (1)  are  2,  2,  —  1,  and  —  1. 
Subtracting  the  sum  of  2,  2,  —  1,  and  —  1  from  —  1,  the  remaining 
root  is  -  3  (§  321). 

EXERCISE  85 

Find  all  the  roots  of  the  following : 

2.  x*  +  6x3-llx2-60x  +  100  =  0. 

3.  9a?  + 105 a*  +  343 x +343  =0, 

4.  4xi  +  32xs  +  63x2-8x-16  =  0. 

5.  af  +  brt-ll  ar3- 49  a2 +  160  a; -100  =  0. 

6.  x4  +  3a?3  +  4a;2  +  3aj  +  l  =  0. 

345.  The  equation  xn  —  a  =  0  can  have  no  multiple  roots ; 
for  the  derivative  of  xn  —  a  is  nxn~l,  and  xn  —  a  and  nx"'1  have 
no  common  factor  except  unity. 

Hence,  the  n  roots  of  xn  =  a  are  all  different- 
It  follows  from  this  that  every  expression  has  two'different 
square  roots,  three  different  cube  roots,  and,  in  general,  n  dif- 
ferent nth.  roots. 


252  ALGEBRA 

LOCATION   OF   ROOTS 

346.  If  two  real  numbers,  a  and  b,  not  roots  of  the 
equation 

^+^tt  !+  .,..  +Pnlv+pn  =  0,  (1) 

when  substituted  for  a?  in  the  first  member,  give  results 
of  opposite  sign,  an  odd  number  of  roots  of  the  equation 
lie  between  a  and  b. 

Let  a  be  algebraically  greater  than  b. 

Let  d,  •  ••,  g  be  the  real  roots  of  (1)  lying  between  a  and  b,  and  h,  •  ••,  &, 
the  remaining  real  roots. 

Let  xn  -f  piX*~l  +  •••  +i?n-]X  +pn  be  denoted  by  X. 
Then,  by  §  318, 

Xr=(x-d)  ...  (x-g)  -(x-h)  .-.  (*-*)  ■  Y;  (2) 

where  F  denotes  the  product  of  the  factors  corresponding  to  the  imagi- 
nary roots,  if  any,  of  (1). 

Substituting  a,  and  then  &,  for  x  in  (2),  the  second  member  becomes 

(a-d)  ...  (a-g)-(a-h)  •••  ((*-*)•  F,  (3) 

and  (p-d)...  (b-g).(b-h)  -  {b-k).Y"'y  (4) 

where  Yf  and  y"  denote  the  values  of  Y  when  sc  is  put  equal  to  a  and  6, 
respectively. 

Since  a  is  greater  than  b,  each  of  the  numbers  d,  •••,  g  is  less  than  a 
and  greater  than  b. 

Whence,  each  of  the  factors  a  —  d,  •••,  a  —  g  is  +,  and  each  of  the 
factors  b  —  d,  •  ■  • ,  b  —  g  is  — . 

Again,  since  none  of  the  numbers  h,  •••,  k  lie  between  a  and  5,  the 
expression  (a  —  K)  •  •  •  (a  —  k)  has  the  same  sign  as  the  expression 

(b-h)  ...  (b-k). 

Also,  y  and  Y"  are  positive ;  for  the  product  of  the  factors  corre- 
sponding to  a  pair  of  conjugate  imaginary  roots  of  (1)  is  positive  for  every 
real  value  of  x  (§  324). 

But  by  hypothesis,  the  expressions  (3)  and  (4)  are  of  opposite  sign. 

Hence,  the  number  of  factors  b  —  d,  •••,  b  —  g -must  be  odd; 
that  is,  an  odd  number  of  roots  lie  between  a  and  b. 

If  the  numbers  substituted  differ  by  unity,  it  is  evident  that  the  inte- 
gral part  of  at  least  one  root  is  known. 


THEORY  OF   EQUATIONS  253 

Ex.     Locate  the  roots  of  or3  -j-  x2  —  6  x  —  7  =  0. 

By  Descartes'  Rule  (§  835),  the  equation  cannot  have  more  than  one 
positive,  nor  more  than  two  negative  roots. 

The  values  of  the  first  member  for  the  values  0,  1,  2,  3,  —  1,  —  2,  and 
—  3  of  x  are  as  follows  : 

J5=0;  -7.  x  =  2;— 7.  a;  =  -  1 ;  —  1.  x  =  -  3  ;  -  7. 

x  =  l;   -11.  x  =  3;  11.  sc  =  -  2  ;  1. 

Since  the  sign  of  the  first  member  is  —  when  x  =  2,  and  -f  when  x  =  3, 
one  root  lies  between  2  and  3. 

The  others  lie  between  —  1  and  —  2,  and  —  2  and  —  3,  respectively. 

In  locating  roots  by  the  above  method,  first  make  trial  of  the  numbers 
0,  1,  2,  etc.,  continuing  the  process  until  the  number  of  positive  roots  de- 
termined is  the  same  as  has  been  previously  indicated  by  Descartes'  Rule. 

Thus,  in  the  above  example,  the  equation  cannot  have  more  than  one 
positive  root ;  and  when  one  has  been  found  to  lie  between  2  and  3,  there 
is  no  need  of  trying  4,  or  any  greater  positive  number. 

The  work  may  sometimes  be  abridged  by  finding  a  superior  limit  to 
the  positive  roots,  and  an  inferior  limit  to  the  negative  roots  of  the  given 
equation  (§§  339,  340),  for  no  number  need  be  tried  which  does  not  fall 
between  these  limits. 

EXERCISE   86 

Locate  the  roots  of  the  following : 
i.   x?  +  4:X2-6  =  0.  5.    x4  +  3x*-±x-l  =  0. 

2.  a?-7x*  +  6x  +  5  =  0.     6.    x4  +  ar*-19  x2-  17  x  +  1  =  0. 

3.  ^  +  3^-7^  + 2  =  0.     7-   x4-4x3  +  6  z-2  =  0. 

4.  f  +  4a;2  +  «-3  =  0.        8.    #4- 7  ^  +  #  +  4  =  0. 

9.    Prove  that  the  equation  x4  —  5x?  —  7  x  —  2  =  0  has  one 
root  between  2  and  3,  and  at  least  one  between  0  and  —1. 

10.  Prove  that  the  equation  x4  —  3  xP  +  x2  —  3  x  —  4  =  0  has 
one  root  between  0  and  —  1,  and  at  least  one  between  3 
and  4. 

11.  Prove  that  the  equation  ar?-r-5a;-f-4  =  0  has  one  root 
between  0  and  —  1. 


254 


alc;ebra 


347.  The  method  of  §  346  is  not  sufficient  to  deal  with 
every  problem  in  location  of  roots. 

Let  it  be  required,  for  example,  to  locate  the  roots  of 

By  §  325,  the  equation  has  at  least  one  real  root. 

By  Descartes'  Rule,  it  has  no  positive  root. 

Putting  x  equal  to  0,  —  1,  —  2,  —  3,  the  corresponding  values 
of  the  first  member  are  1,  1,  1,  and  —  5,  respectively. 

Then,  the  equation  has  either  one  root  or  three  roots  between 
—  2  and  —  3 ;  but  the  methods  already  given  are  not  sufficient 
to  determine  which. 

Sturm's  Theorem  (§  350)  affords  a  method  for  determining 
completely  the  number  and  situation  of  the  real  roots  of  an 
equation. 

It  is  more  difficult  of  application  than  the  method  of  §  346, 
and  should  be  used  only  in  cases  which  the  latter  cannot 
resolve. 

348.  Graphical  Representation. 

The  graph  of  an  expression  of  higher  degree  than  the  sec- 
ond, with  one  unknown  number,  may  be  found  as  in  §  51. 


Ex.     Find  the  graph  of 

[f(x)  .        ,  x>-  2  x2  -2  z  +  3. 

Put/O)  =  z3  -  2  x2  -  2  x  +  3. 

If  3=0, /(a)  =3. 

If  x  =  l,/(x)  =  0. 
Ifx  =  2,/(x)  =  -l. 
If  x  =  -2,/(x)  =  -9. 
Ifx=-l,/(x)=2. 
If  x  =  3,/(x)  =6. 
etc. 

The  graph  is  the  curve  ABC,  which  extends  in  either  direction  to  an 
indefinitely  great  distance  from  XX. 


THEORY  OF   EQUATIONS  255 

349.  Graphical  Location  of  Roots. 

The  principle  of  §  220  holds  for  the  graph  of  the  first  mem- 
ber of  an  equation  of  higher  degree  than  the  second,  with  one 
unknown  number. 

Thus,  the  graph  of  §  348  intersects  XX'  at  x  =  1,  between 
x  =  2  and  x  =  3,  and  between  x  =  —  1  and  x  =  —  2. 

And  the  equation  Xs  —  2  x2  —  2^+3=0  has  one  root  equal 
to  1,  one  between  2  and  3,  and  one  between  —  1  and  —  2. 

This  may  be  verified  by  solving  the  equation ;  the  factors  of  the  first 
member  are  x  —  1  and  x2  —  x  —  3. 

This  method  of  locating  roots  is  simply  a  graphical  repre- 
sentation of  the  process  of  §  346,  and  is  subject  to  the  limita- 
tions stated  in  §  347. 

If  the  graph  is  tangent  to  XX',  the  equation  has  two  or 
more  equal  roots  (compare  §  220,  Fig.  2) ;  if  it  does  not  inter- 
sect XX',  the  equation  has  no  real  root. 

The  note  to  the  example  of  §  346  applies  with  equal  force  to  the 
graphical  method  of  locating  roots. 

EXERCISE   87 

Locate  the  roots  of  the  following  graphically : 

i.   a^ -3*  — 1  =  0.  4-   x3- 8  ^  +  19  a -12  =  0. 

2.  a4  +  2z2  +  3  =  0.  5.    aj8  +  7aj24-14a?  +  8  =  0. 

3.  ^-7ar9  +  12a-5  =  0.  6.   x'-Sx2-  2  x  +  5  =  0. 

350.  Sturm's  Theorem. 

Let  a^+jPi^-f  •••  -fiV-i#+P»  =  0  (1) 

be  an  equation  from  which  the  multiple  roots  have  been  re- 
moved (§  343). 

Let  xn -{- p^91"1  +  •••  +pn~lx+pn  be  denoted  by  f(x),  and  let 
f(x)  denote  the  first  derivative  of  f(x)  (§  341). 

Dividing  f(x)  by  f(x),  we  shall  obtain  a  quotient  #i>  with 
a  remainder  of  a  degree  lower  than  that  of  fi(x). 

Denote   this  remainder,  ivith  the   sign  of  each  of  its   terms 


256  alokbka 

changed,  by  f2(x)9  and  divide  f(x)  by  f(x)9  and  so  on ;  the 
operation  being  precisely  the  same  as  that  of  finding  the 
H.  C.  F.  of  f(x)  and  fi(x)9  except  that  the  signs  of  the  terms 
of  each  remainder  are  to  be  changed,  while  no  other  changes 
of  sign  are  permissible.  ' 

Since,  by  hypothesis,  fix)  =  0  has  no  multiple  roots,  f(x) 
and  fi(x)  have  no  common  divisor  except  1  (§  343) ;  and  we 
shall  finally  obtain  a  remainder/,^)  independent  of  x. 

The  expressions  f(x),f1(x)9f2(x),  •••,/„(»),  are  called  Sturm's 
Functions. 

The  successive  operations  are  represented  as  follows : 

f(x)  =  Qlf(x)-f2(x)9  (2) 

fx(p)  =  Q2f2(x)  -f(x)9  (3) 

Mx)  =  Qsf*(x)-fi(x),  (4) 


fn-2(x)  =  Qn_}fn.1(x)-fn(x). 

We  may  now  enunciate  Sturm's  Theorem : 

Let  two  real  numbers,  a  and  b,  be  substituted  in  place 
of  x  in  Sturm's  Functions,  and  the  signs  noted. 

The  difference  between  the  number  of  variations  of 
sign  (§  334)  in  the  first  case  and  that  in  the  second  is 
equal  to  the  number  of  real  roots  of  /(as)  =0  lying  be- 
tween a  and  b. 

The  proof  of  the  theorem  depends  upon  the  following 
principles : 

I.  Tivo  consecutive  functions  cannot  both  become  0  for  the 
same  value  of  x. 

For  if,  for  any  value  of  x9f(x)—0  and  f2(x)  =  0,  then  by 
(3),  ./<$(#)  =  0 ;  and  since  f2(x)  =  0  and  f(x)  =  0,  by  (4)  f4(x)  =  0 ; 
continuing  in  this  way,  we  shall  finally  have/„(#)  =  0. 

But  by  hypothesis,  fn(x)  is  independent  of  x9  and  conse- 
quently cannot  become  0  for  any  value  of  x. 

Hence,  no  two  consecutive  functions  can  become  0  for  the 
same  value  of  x. 


THEORY  OF   EQUATIONS  257 

II.  If  any  function,  except  f(x)  and  fn(x),  becomes  0  for  any 
value  of  x,  the  adjacent  functions  have  opposite  signs  for  this 
value  of  x. 

For  if,  for  any  value  of  x,  f2(x)  =  0,  then,  by  (3),  we  must 
have  f(x)  =  —f3(x)  for  this  value  of  x. 

Therefore,/^)  and  f(x)  must  have  opposite  signs  for  this 
value  of  x ;  for,  by  I,  neither  of  them  can  equal  zero. 

III.  Let  c  be  a  root  of  the  equation  f(x)  =  0,  where  f(x) 
is  any  function  except  f(x)  aj\o\fn(x). 

By  Ilyf^x)  &ndfr+1(x)  have  opposite  signs  when  x=  c. 

Let  h  be  a  positive  number,  so  taken  that  no  root  of  fr-i(x) 
=  0,  or  fr+i(x)  =  0  lies  between  c  —  h  and  c  +  h. 

Then,  as  x  changes  from  c—h  to  c  +  k,  no  change  of  sign 
takes  place  in  fr_i(x),  or  fr+i(x)  ;  while  fr(x)  reduces  to  zero, 
and  changes  or  retains  its  sign  according  as  the  root  c  occurs 
an  odd  or  even  number  of  times  in  f(x)  =  0. 

Therefore,  for  values  of  x  between  c  —  h  and  c,  and  also 
for  values  of  x  between  c  and  c  +  h,  the  three  functions 
f-i(x)>  fXx)>  an(i  fr+i(x)  present  one  permanence  and  one 
variation.     * 

Hence,  as  x  increases  from  c—h  to  c  +  h,  no  change  occurs  • 
in  the  number  of  variations  in  the  functions  f_i(x),  fr(x),  and 
fr+i(x) ;  that  is,  no  change  occurs  in  the  number  of  variations 
as  x  increases  through  a  root  of  fr(x)  =  0. 

IV.  Let  c  be  a  root  of  the  equation  f(x)  =  0 ;  and  let  h  be  a 
positive  number  so  taken  that  no  root  of  f(x)  =  0  lies  between 
c  —  h  and  c  +  h. 

Then  as  x  increases  from  c  —  h  to  c  -f  h,  no  change  of  sign 
takes  place  in  f(x),  while  f(x)  reduces  to  zero,  and  changes 
sign. 

Now  if  we  put  x  =  c  —  h  in  (1),  the  first  member  becomes 

Expanding  the  powers  of  c  —  h  by  the  Binomial  Theorem, 


258  ALGEBRA 

and   collecting    the    terms    involving    like   powers    of   h,   we 

have  . 

c1i+plcn  *  +  •••  +pn-ic+pn 

-hlnc^  +  in-l^c"  2  +  •••  +pn-i] 

+  terms  involving  h2,  h3,  •  ••,  hn.  (5) 

But  since  c  is  a  root  of  f(x)  =  0,  we  have  by  (1), 

Also,  it  is  evident  that  the  coefficient  of  —  h  is  the  value  of 
fx(x)  when  c  is  substituted  in  place  of  x;  let  this  be  denoted  by 
A ;  then  (5)  reduces  to 

—  JiA  -f  terms  involving  7i2,  7i3,  •  • •,  hn.  (6) 

In  like  manner,  the  value  of  f(x)  when  x  is  put  equal  to 
c  +  h,  is 

+  7^yl  +  terms  involving  h2,  7is,  •  •  •  hn.  (7) 

Now,  if  7i  be  taken  sufficiently  small,  the  signs  of  the  ex- 
pressions (6)  and  (7)  will  .be  the  same  as  the  signs  of  their 
first  terms,  —  h A  and  -f  hA,  respectively. 

Hence,  if  h  be  taken  sufficiently  small,  the  sign  of  (6)  will 
be  contrary  to  the  sign  of  A,  and  the  sign  of  (7)  will  be  the 
same  as  the  sign  of  A. 

Therefore,  for  values  of  x  between  c  —  h  and  c,  the  functions 
/  (x)  and  fi(x)  present  a  variation,  and  for  values  of  x  between 
c  and  c  +  h  they  present  a  permanence. 

Hence,  a  variation  is  lost  as  x  increases  through  a  root  of 
the  equation  f(x)  =  0. 

We  may  now  prove  Sturm's  Theorem ;  for  as  x  increases 
from  b  to  a,  supposing  a  algebraically  greater  than  b,  a  varia- 
tion is  lost  each  time  that  x  passes  through  a  root  of  f(x)  =  0, 
and  only  then;  for  when  x  passes  through  a  root  of  fr(x)  =0, 
where  fr(x)  is  any  function  except  f(x)  and  /»(#),  no  change 
occurs  in  the  num  ber  of  variations. 

Hence,  the  number  of  variations  lost  as  x  increases  from  b 
to  a  is  equal  to  the  number  of  real  roots  of  Ar=0  included 
between  a  and  b. 


THEORY  OF   EQUATIONS  259 

351.  It  is  customary,  in  applications  of  Sturm's  Theorem, 
to  speak  of  the  substitution  of  an  indefinitely  great  positive 
number  for  x,  in  an  expression,  as  substituting  -f-  co  for  x ;  and 
the  substitution  of  a  negative  number  of  indefinitely  great 
absolute  value  as  substituting  —  oo  for  x. 

The  substitution  of  +  oo  and  —  go  for  x  in  Sturm's  Func- 
tions determines  the  number  of  real  roots  of  f(x)  —  0. 

The  substitution  of  +  oo  and  0  for  x  determines  the  number 
of  positive  real  roots,  and  the  substitution  of  —  oo  and  0  the 
number  of  negative  real  roots. 

Since  Sturm's  Theorem  determines  the  number  of  real  roots 
of  an  equation,  the  number  of  imaginary  roots  also  becomes 
known  (§  316). 

352.  If  a  sufficiently  great  number  be  substituted  in  place 
of  x  in  the  expression 

f(x)=pdxn+plxn~1  +  .-.  +Pn_lX+pn, 

the  sign  of  the  result  will  be  the  same  as  the  sign  of  its  first 
term,  p0xn. 

It  follows  from  the  above  that : 

If  +  oo  be  substituted  in  place  of  oc  in /(a?),  the  sign  of 
the  result  will  be  the  same  as  the  sign  of  its  first  term. 

If  -oo  be  substituted  in  place  of  as  in /(as),  the  sign  of 
the  result  will  be  the  same  as,  or  contrary  to,  the  sign  of 
the  first  term,  according  as  the  degree  of  /  (as)  is  even  or 
odd. 

353.  Examples. 

i.   Determine  the  number  and  situation  of  the  real  roots  of 

aj3_2ar-x4-l=0. 

Here,  f(x)  =  x* -2  oc?  -  x +  \,  and  fx(x)  =  3  x2  -  4  x  -  1. 

In  the  process  of  finding /2(x), /3(x),  etc.,  any  positive  numerical 
factors  may  be  omitted  or  introduced  at  pleasure,  for  the  sign  of  the 
result  is  not  affected  thereby  ;  in  this  way  fractions  may  be  avoided. 


260  ALGEBRA 

In  the  present  case,  we  multiply  /(x)  by  3,  to  make  its  first  term 
divisible  by  3  x2. 

3x3-4x2-      x 
-2x2-    2x  +  3 

3 

-6a;2-   6x  +  9(-2 
—  6x2  +  8x  +  2 
7)-14x+7 

-  2  x  +  1         Then,  /2(x)  =  2  x  -  1. 
3X2_  4x_i 

2 


2x-  1)6 x2-  8x-2(3x 
6x2-  3x 
-  6x-2 

2 


_10x-4(-6 
—  10  x  +  5 

~^9         Then,/3(x)=9. 

Substituting  —  oo  for  x  in  /(x),  /i(x),  /2(x),  and  /s(x),  the  signs  are 
— ,  +  ,  — ,  and  +  ,  respectively  (§  352);  substituting  0  for  x,  the  signs 
are  +,  — ,  — ,  +,  respectively  ;  and  substituting  +  oo  for  x,  the  signs  are 
all  +  . 

Hence,  the  roots  of  the  equation  are  all  real,  and  two  of  them  are  posi- 
tive and  the  other  negative  (§  351). 

We  now  substitute  various  numbers  to  determine  the  situation  of  the 
roots : 


/w 

AW 

Mx) 

/•(*) 

x—  — 

GO, 

- 

+ 

— 

+        * 

3  variations. 

x=  — 

1, 

— 

+ 

- 

+ 

3  variations. 

x  =  0, 

+ 

- 

— 

+ 

2  variations. 

X  =  1, 

- 

— 

+ 

+ 

1  variation. 

35  =  2, 

— 

+ 

+ 

+ 

1  variation. 

x  =  3, 

+ 

+ 

+ 

+ 

no  variation. 

X  =  00 

» 

+ 

+ 

+ 

+ 

no  variation. 

We  then  know  that  the  equation  has  one  root  between  0  and  —  1,  one 
between  0  and  1,  and  one  between  2  and  3. 

2.    Determine  the  number  and  situation  of  the  real  roots  of 

4  x3  —  6  x  —  5  =  0. 


THEORY   OF   EQUATIONS  261 

'  Here,  /(x)  =  4x3  —  6x— 5;  and  fx(x)  =  12  x2  —  0,  or  2  x2  -  1,  omitting 
the  factor  6. 

•      2x2-l)4x3-6x-5(2x 

4  x3  -  2  x 


-4x-5 

2  x2  -     1 
2 

Then,/2(x)  =  4x  +  5. 

4x  +  5)4x2-     2(x 
4  x2  +     5  x 

—     5x—   2 
4 

-20x-  8(- 
-  20  x  -  25 
17 

-5 

Then,  /3(x)  =  -  17. 

The  last  step  in  the  division  may  be  omitted;  for  we  only  need  to 
know  the  sign  of  /s(x),  and  it  is  evident  by  inspection,  when  the  re- 
mainder —  5  x  —  2  is  obtained,  that  the  sign  of /3(x)  will  be  — . 


/(*) 

/iW 

h(x) 

Mx) 

X  =  —  QO  , 

— 

+ 

— 

— 

2  variations. 

x  =  0, 

— 

- 

+ 

— 

2  variations. 

x  =  l, 

— 

+ 

+ 

- 

2  variations. 

SB  =  2, 

+ 

+ 

+ 

- 

1  variation. 

X  =  GO  , 

+ 

+ 

+ 

— 

1  variation. 

Therefore,  the  equation  has  a  real  root  between  1  and  2,  and  two 
imaginary  roots. 

In  substituting  the  numbers,  it  is  best  to  work  from  0  in  either  direc- 
tion, stopping  when  the  number  of  variations  is  the  same  as  has  been 
previously  found  for  +  oo  or  —  co  ,  as  the  case  may  be. 

EXERCISE  88 

Determine  the  number  and  situation  of  the  real  roots  of : 
i.    xs  +  2x2-x-l  =  0.  5-   x*-4:X2  +  x+3  =  0. 

2.  xs  +  3x-5  =  0.  6.   a4-8a2-8#  +  l  =  0. 

3.  x3-5x2+2x  +  6  =  0.  7-   a4  +  2  ar*- 5ar-10a-3=0. 

4.  xs  +  x2-15x-28  =  0.        8.   a4-r-3ar*-3z  +  l  =  0. 


262  ALGEBRA 

XV.    SOLUTION  OF  HIGHER  EQUATIONS 

354.  Synthetic  Division  (§  107)  not  only  abbreviates  the 
process  of  division,  but  its  application  is  of  importance  in  the 
solution  of  many  forms  of  higher  equations  containing  either 
commensurable  or  incommensurable  roots. 

COMMENSURABLE   ROOTS 

355.  We  use  the  term  commensurable  root,  in  Chapter  XV, 
to  signify  a  rational  root  expressed  in  Arabic  numerals. 

356.  By  §  322,  an  equation  of  the  nth  degree  in  the  general 
form  (§  312),  with  integral  numerical  coefficients,  cannot  have 
as  a  root  a  rational  fraction  in  its  lowest  terms. 

Therefore,  to  find  all  the  commensurable  roots  of  such  an 
equation,  we  have  only  to  find  all  its  integral  roots. 

Again,  by  §  320,  the  last  term  of  an  equation  of  the  above 
form  is  divisible  by  every  integral  root. 

Hence,  to  find  all  the  commensurable  roots,  we  have  only  to 
ascertain  by  trial  which  integral  divisors  of  the  last  term  are  roots 
of  the  equation. 

The  trial  may  be  made  in  two  ways : 

I.  By  substitution  of  the  supposed  root. 

II.  By  dividing  the  first  member  of  the  equation  by  the 
unknown  number  minus  the  supposed  root  (§  315). 

In  this  case,  the  operation  may  be  conveniently  performed 
by  Synthetic  Division  (§  107). 

In  the  case  of  small  numbers,  such  as  ±1,  the  first  method 
may  be  the  most  convenient. 

The  second  has  the  advantage  that,  when  a  root  has  been 
found,  the  process  gives  at  once  the  depressed  equation  (§  317) 
for  obtaining  the  remaining  roots. 

Work  may  sometimes  be  saved  by  finding  a  superior  limit  to  the 


SOLUTION   OF    HIGHER  EQUATIONS  263 

positive,  and  an  inferior  limit  to  the  negative,  roots  (§§  339,  340); 
for  no  number  need  be  tried  which  does  not  fall  between  these 
limits. 

Descartes'  Rule  of  Signs  (§  335)  may  also  be  advantageously  employed 
to  shorten  the  process. 

Any  multiple  root  should  be  removed  (§  343)  before  applying  either 
method. 


Ex.   Find  all  the  roots  of  x4  - 15  x2  + 10  x  +  24  =  0. 

By  Descartes'  Rule,  the  equation  cannot  have  more  than  two  positive 
roots. 

Changing  the  signs  of  the  alternate  terms  commencing  with  the  second, 
we  have  z4  -  15  x2  —  10  x  +  24  =  0. 

Then,  the  given  equation  cannot  have  more  than  two  negative  roots 
(§  336). 

The  integral  divisors  of  24  are  ±1,  ±2,  ±3,  ±4,  ±6,  ±8,  ±12,  and 
±24. 

By  substitution,  we  find  that  1  is  not,  and  that  —  1  is,  a  root  of  the 
equation. 

Dividing  the  first  member  by  x  —  2,  x  —  3,  etc.,  by  the  method  ex- 
plained in  §  108,  we  have 

1  +  o  -  15  +  10  +  24  j_2  1  +  0  -  15  +  10  +  24  [3 

2  4-22-24  3  9-18-24 

2-11-12,        0  Rem.  3  -  6  -    8,       0  Rem. 

The  work  shows  that  2  and  3  are  roots  of  the  given  equation ;  and 
since  the  equation  cannot  have  more  than  two  positive  roots,  these  are 
the  only  positive  roots. 

The  remaining  root  may  be  found  by  dividing  24  by  the  product  of 
!  —  1,  2,  and  3  (§  321),  or  by  the  same  process  as  above. 

Dividing  the  first  member  by  x  +  2,  x  +  3,  etc.,  we  have 

1  +  0-15+10+24  [  -2  1  +  0-15+10+24  [^-3 

-2  4       22  -  64  -3  9       18-84 

-2-11       32-40  -3-    6       28-60 

1  +o_  15  +10  +241-4 
-4       16  -    4-24 
-4         16         0 

The  work  shows  that  the  remaining  root  is  —  4. 


264  ALGEBRA 

357.  By  §  829,  an  equation  of  the  nth  degree  in  the  general 
form,  with  fractional  coefficients,  may  be  transformed  into 
another  whose  coefficients  are  integral,  that  of  the  first  term 
being  unity. 

The  commensurable  roots  of  the  transformed  equation  may 
then  be  found  as  in  §  356. 

*  Ex.   Find  all  the  roots  of  4  a?  -  12  x2  +  27  x  - 19  =  0. 
Dividing  through  by  the  coefficient  of  x8,  we  have 

4         4 

Proceeding  as  in  §  329,  it  is  evident  by  inspection  that  the  multiplier  2 
will  remove  the  fractional  coefficients  ;  the  transformed  equation  is 

X3_  2  .  3  x2  +  22 .21*  -23  •  -  =  0, 
4  4 

or,  Xs  -  6  x2  +  27  x  -  38  =  0  ;  (1) 

whose  roots  are  those  of  the  given  equation  multiplied  by  2. 
By  Descartes'  Rule,  equation  (1)  has  no  negative  root. 
The  positive  integral  divisors  of  38  are  1,  2,  19,  and  38. 
Dividing  the  first  member  by  x  —  1,  x  —  2,  etc.,  we  have 

1  _  6  +  27  -  38  |_1_  1  -  6  +  27  -  38  |_2 

1-5      22  2-8      38 

-  5      22  - 16  -  4      19        0 

The  work  shows  that  2  is  a  root  of  (1). 

The  remaining  roots  may  now  be  found  by  depressing  the  equation; 
it  is  evident  from  the  right-hand  operation  above  that  the  depressed  equa- 
tion is  x2  —  4  x  +  19  =  0. 

Solving  by  the  rules  for  quadratics,  x  =  2  ±  V—  15. 

Then,  the  three  roots  of  (1)  are  2  and  2  ±  V—  15. 

Dividing  by  2,  the  roots  of  the  given  equation  are  1  and  1  ±  V—  15. 

EXERCISE  89 

Find  all  the  commensurable  roots  of  the  following,  and  the 
remaining  roots  when  possible  by  methods  already  given. 

i.    ,rs-9x2  +  23x-15  =  Q.         3.   aj8  +  12a?2  +  44a?  +  48  =  0. 

2.   aj3_8x2-h   5a>  +  14  =  0.         4.   x*  +  Ax2-  9»-36  =  0. 


SOLUTION   OF   HIGHER   EQUATIONS       '       265 

5.  3ar3H-4ar>-13a;-f 6  =  0. 

6.  4a8+16a2-7a>-39=0. 

7.  a4+10ar3  +  3ox2  +  o0z  +  24  =  0. 

8.  a4- 5  &»  + 20  a?- 16  =  0. 

9.  a*-15a*  +  65a?-105x  +  5±  =  0. 

10.  aj4  +  8a^4-ll^2-32a!-60  =  0. 

11.  xi-2x*-l7x2  +  18x  +  72  =  0. 

12.  4a4-12a8-9ar9  +  47x-30  =  0. 

13.  6a*4-7a8-37a2  +  8a;4-12  =  0. 

14.  ar5  +  8a;4-7ar3-103a2  +  69x  +  18  =  0. 

15.  3a4-f-2ar5-18x2  +  8  =  0. 

16.  x4  +  #  -  6  x2  + 16  a?  -  32  =  0. 

♦  RECIPROCAL   OR  RECURRING   EQUATIONS 

358.  A  Reciprocal  Equation  is  one  such  that  if  any  number 
is  a  root  of  the  equation,  its  reciprocal  is  also  a  root. 

It  follows  from  the  above  that,  if  -  be  substituted  for  x  in 

x 

a  reciprocal  equation,  the  transformed  equation  will  have  the 

same  roots  as  the  given  equation. 

359.  Let 

xn+plxn~1+p^n-2+  .-  +pH-jx?+pn-iX+pH  =  0         (1) 

be  a  reciprocal  equation. 

Putting  -  for  x,  the  equation  becomes 
x 

x11     xn~x      x"-1  X1  X 

Clearing  of  fractions,  and  reversing  the  order  of  the  terms, 

PnXn  +Pn-\Xn~1  +i>n-2#n~2  4"  '—  "f  P»X2  +  Pl%  +  1=0. 


266        •  ALGEBRA 

Dividing  through  by  pn, 

xn  +Pj}=_\Xn-l  +P_rL=2xn-2  +    ...    +  P?^  +  Si  X  +  -I  =  0.  (2) 

Pn  Pn  Pn  Pn  Pn 

By  §  358,  this  equation  has  the  same  roots  as  (1)  ;  and  hence  the  fol- 
lowing relations  must  hold  between  the  coefficients  of  (1)  and  (2), 

p1=^,p2=^,..-,Pn-2=^,Pn-i=P-±,pn  =  ±-.  (3) 

Pn  Pn  Pn  Pn  Pn 

From  the  last  equation,  pn%  =  1  ;  whence,  pn  =  ±  1. 
Substituting  the  value  of  pn,  the  equations  (3)  become 

Pi  -   ±  Pn-l,    P2  =    ±  Pn-2,  "•  5 

all  the  upper  signs,  or  all  the  lower  signs,  being  taken  together. 

We  then  have  four  varieties  of  reciprocal  equations : 

1.  Degree  odd,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute  value 
and  of  like  sign  ;  as,  x3  —  2  x2  —  2  x  + 1  =  0. 

2.  Degree  odd,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute  value 
and  of  opposite  sign ;  as,  3  x5  -f  2  x4  —  Xs  +  x2  —  2  x  —  3  =  0. 

3.  Degree  even,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute  value 
and  of  like  sign ;  as,  x*  —  5  Xs  +  6  x2  —  5  x  -f- 1  =  0. 

4.  Degree  even,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute  value 
and  of  opposite  sign,  and  middle  term  wanting ;  as, 

2^6  +  3aj5-7^4  +  7^2-3a?-2  =  0. 

On  account  of  the  properties  stated  above,  reciprocal  equa- 
tions are  also  called  Recurving  Equations, 

360.  Every  reciprocal  equation  of  the  first  variety  may  be 
written  in  the  form 


SOLUTION   OF   HIGHER   EQUATIONS  267 

or,  i)0(^  +  l)+i>i(^-1+^)-hp2(^-2  +  ^)+  •••  =0;    (1) 

or,         p0(xn  +  1)  +p&(&~*  +  !)  +P^(xn~4  + 1)  +  ...  =  0 ; 
the  number  of  terms  being  even. 

Since  n  is  odd,  each  of  the  expressions  xn  +  1,  xn~2  +  1,  etc.,  is  divisible 
by  x+  1. 

Therefore,  —  1  is  a  root  of  the  equation  (§  315). 

Dividing  the  first  member  of  (1)  by  x  +  1,  the  depressed  equation  is 

PoOw_1  —  xn~2  +  xn~s  —  •••  +  x2  —  X  +  1) 
+  Pl(Xn~2  —  Xn"3  +  £n~4  —   •••   +XS  -  x2  +  x) 

+  p2(xn~s  -  xn~*  +  xn~5  -  •••  +  x4  —  Xs  +  z2)  +  ...  =  0. 
Or,  jpo^n_1  +  (Pi  -  Po)£n~2  +  (P2-.P1  +Po)xn~*  +  ... 

+  (JP2  -  Pi  +  Po)x2  +  (Pi-po)x+p0  =  0; 

which  is  a  reciprocal  equation  of  the  third  variety. 

361.   Every  reciprocal  equation  of  the  second  variety  may 
be  written  in  the  form 

PoXn  -f  pjX**1  +  p2xn~2  +  •  •  •  —  peps2  —  plx—p0  =  0, 

or,  p0(xn  -  1)  -f  p^x71-1  -  x)  +  p2(ajn"2  -  #2)  +  •  •  •  =  0,     (1) 

or,        Po(xn  —  1)  +  p&ix71-2  —  1)  +  p2x2(xn~4  —  1)  +  •  •  •  =0. 

Since  each   of  the   expressions  xn  —  1,  xn~2  —  1,  etc.,  is  divisible  by 
x  —  1,    +  1  is  a  root  of  the  equation. 

Dividing  the  first  member  of  (1)  by  x  —  1,  the  depressed  equation  is 

Po(xn~l  +  xn~2  H-  xn~3  +  ...  +  x2  +  x  +  1) 

+  PiOcn~2  +  xn~3  +x"~4  -f  ..."  +  xs  +  a*2  +  a) 

+  P20n-3  +  £"~4+  xn-5+  ...  +x4  4-a^  +  x2)  +  •••  =0, 

or,  poo71-1  +  (Pi  +  Po)^n~2  +  (pa  -f  Pi  +Po)xn~s  +  ••• 

+  (Pi  +  Pi  +  Po)x2  +  (p  -f  p0).*  +  Po  =  0  ; 

which  is  a  reciprocal  equation  of  the  third  variety. 


268  ALGEBRA 

362.  Every  reciprocal  equation  of  the  fourth  variety  may 
be  written  in  the  form 

p0(xn  -1)+  PiO""1  -  *)  4-i>2(^"2  -  a8)  4-  •  •  •  =  0,     (1) 
or,        p0(xn  -  1)  +PxX{ar-*  -  1)  +p&\xn~*  -  1)  4-  •  •  •  =  0 ; 
the  number  of  terms  being  even  (§  359). 

Since  each  of  the  expressions  xn  —  1,  xn~2  —  1,  etc.,  is  divisible  by 
x2  —  1,  both  1  and  —  1  are  roots  of  the  equation. 

Dividing  the  first  member  of  (1)  by  x2  —  1,  the  depressed  equation  is 

p0(xn~2  +  xn~4  +  xn~&  +  •  ••  4  z4  4  x2  4- 1) 

+  lh(xw-8  4  £TC-5  4  xn"7  + h  x5  4  «3  +  £) 

4  P2(xn~4  4  £w_6  4  xn~8  4-  ...  4  x6  4  x4  4  x2)  +  ••  •  =  0, 
or,  i>0^n-2  +  piz*-3  4  (i>2  4  Po)xTO"4  4  — 

+  (p2  4p0)^2  +W»  +  Po  =  °  ; 
which  is  a  reciprocal  of  the  third  variety. 

363.  Every  reciprocal  equation  of  the  third  variety 
may  be  reduced  to  an  equation  of  half  its  degree. 

Let  the  equation  be 

i>o^m4-Pi^m-14- r-pm_2^m+2+i>m-i^w+1+i>m^ 

+Pm-ixm-1  4-i>m_2^w"2+  •••  +PiX  +pQ  =  0. 
Dividing  each  term  by  xm,  the  equation  may  be  written 

Po[xm+  ±-)+Pl(xm-l  +  -^—)+  ... 
V         x™)  \  xm-^j 

+pm-2(x2  4  i]  +*-i(*  +£)  +P-  =  0.  (1) 

Put     x+-=y. 

x 


Then,  a;2  4  ~  =  fa  +-V-  2  =  y2-  2  ; 
x2      \       x] 


=  y(y2-2)-y  =  y3-3y; 


SOLUTION   or   HIGHER  EQUATIONS  269 

=  y(ys  -  3  y)-  (y2  -  2)  =  y*  -  4  y*  +  2  ;  etc. 
In  general, 

x'-     V       &/\  z'-1/      \  xr-2J 

an  expression  of  the  rth  degree  with  respect  to  y. 

Substituting  these  values  in  (1),  the  equation  takes  the  form 

wm  +  <xym~Y  +  &ym~2  +  •••  =  o. 

364.  It  follows  from  §§  360  to  363  that  any  reciprocal  equa- 
tion of  the  degree  2  m  -f  1,  and  any  reciprocal  equation  of  the 
fourth  variety  of  the  degree  2  m  +  2,  can  be  reduced  to  an 
equation  of  the  mth  degree. 

365.  Ex.    Solve  2  x5-5xi-13xi +  13X2 +  5x-  2=0. 

The  equation  being  of  the  second  variety,  one  root  is  1  (§  361). 
Dividing  by  x  —  1,  the  depressed  equation  is 

2x4-3x3-16x2-3x  +  2  =  0; 

a  reciprocal  equation  of  the  third  variety. 

Dividing  by  x2,     2  (x2  -f  -)  -  3  (x  +  -\  -  16  =  0. 

Putting  x  + 1  =  y,  and  x2  +  —  =  y1  -  2  (§  363),  we  have 
x  x2 

20/2-2)-  3  y  -16  =  0. 

Solving  this  equation,  ?/  =  4  or  ~  f . 

Taking  the  first  value,     x  +  -  =  4,  or  x2  —  4  x  +  1  =0. 
x 

Whence,  x  =  2  ±  V3. 

1  5 

Taking  the  second  value,  x  +  -= ,  or  2  x2  +  5  x  +  2  =  0. 

x  2 

Whence,  x  =  —  2  or  —  £. 

The  roots  of  the  given  equation  are  1,  —  2,  —  £,  and  2  ±  V&. 
That  2  -f  V3  and  2  —  V3  are  reciprocals  may  be  shown  by  multiplying 
them  ;  thus,  (2  +  V3)(2  -  \/3)  =4-3  =  1. 


270  ALGEBRA 

EXERCISE  90 
Solve  the  following  equations  : 
i.   4aj3  +  21arJ  +  21a  +  4  =  0.      3.   Xs  -  5  x2  -  5x  +  l  =  0. 
2.   xi  +  4x2-±x-l  =  0.  4.    6a4 -P  13a8- 13  a- 6  =  0. 

5.  24a4-10ar3-77a2-10a  +  24  =  0. 

6.  x"  +  2x4-5x3  +  5x2-2x-l  =  0. 

7.  5  ar5  -  56  a4  +  131  a?  +  131a2-  56  x  +  5  =  0. 

8.  3ar5  +  4a4-23#3-23a2+4a;  +  3=0. 

9.  6ar5-7z4-27a3+27a2  +  7a;-6=0. 

10.   10x6-19^-19x4  +  19^  +  19aj-10  =  0. 

366.  Binomial  Equations. 

A  Binomial  Equation  is  an  equation  of  the  form  xn=  a. 
Binomial   equations   are   also  reciprocal  equations,  and,  in 
certain  cases,  may  be  solved  by  the  method  of  §  365. 

EXERCISE  91 

Solve  the  following  equations : 

1.   x5  =  1.  2.   x*  =  —  1.  3.    ar5  =  a5.    (Put  x  =  ay.) 

CUBIC   EQUATIONS 

367.  A  Cubic   Equation  is  an  equation  of   the  third   degree 
containing  but  one  unknown  number. 

368.  By  §  333,  the  cubic  equation 

ar3  +p$P  +  p2x  +ps  =  0, 

where  pY  is  not  zero,  may  be  transformed  into  another  whose 

second  term  shall  be  wanting  by  substituting  y  —  ~  for  x. 

o 

Hence,  every  cubic  equation  can  be  reduced  to  the  form 
x]  +  ax  +  6  =  0. 


SOLUTION   OF   HIGHER    EQUATIONS  271 

369.    Cardan's  Method  for  the  Solution  of  Cubics. 

Let  it  be  required  to  solve  the  equation  Xs  +  ax  -+-  b  =  0. 
Putting  x  =  y  +  z,  the  equation  becomes 

y3  +  3yz(y  +  z)+z3  +  a(y  +  z)+b  =  0, 

or,  y8  +  »8  +  (3^  +  a)(y  +  *)  +  &  =  0. 

We  may  give  such  a  value  to  z  that  3yz  +  a  shall  equal  zero. 

Whence,  z  =  —  -—  •  (1) 

3y 

Then,  $*+**+*  =0.  (2) 

Substituting  the  value  of  z  from  (1)  in  (2),  we  have 

2/3-  — +  6=0,  or  ys  +  bif-  —  =  0. 

This  is   an   equation  in  the  quadratic  form  (Exercise  44, 
Note  3). 

Solving  by  the  rules  for  quadratics,  we  have 


*'=-h\f+fr  p 


Thenbj(2),    2<  =  -s.-S  =  -|T^  +  |.  (4) 

Taking  the  upper  signs  before  the  radical  signs,  in  (3)  and 
(4),  and  substituting  in  the  equation  x  =  y  +  z,  we  have 


The  lower  signs  before  the  radical  signs  give  the  same  value  of  x. 

The  other  two  roots  may  be  found  by  depressing  the  given 
equation  (§  317). 

Ex.    Solve  the  equation  x3  +  3X2  —  6  a?  -f-  20  =  0.        ^ 

We  first  transform  the  equation  into  another  whose  second  term  shall 
be  wanting. 


272  ALGEBRA 

Putting  x  =  y-%  =  y-l(§  368),  we  have 

2,3  _  3y2  +  sy  -  1  +  3?/2  _  6y  +  3  -  Gy  +  6  +  20  =  0, 
or  ?/3  -  9  y  +  28  =  0. 

To  solve  the  latter  equation,  we  substitute  a  =  —  9  and  b  =  28  in  (5), 
§  369.  

Thus.  y  =3  V-  14  +  v/196  -  27  +  \/-  14  -  Vl96  -  27 

=  v^Ti  +  ^T27  =  -  1  -  3  =-  4. 

Therefore,         x  =  y  —  1  =  —  5. 

Dividing  the  first  member  of  the  given  equation  by  aj  -f  5,  the  depressed 
equation  is  <*-2z  +  4  =  0. 

Solving,  x  =  1  ±  V—  3. 

Thus,  the  roots  of  the  given  equation  are  —  5  and  1  ±  V—  3. 

EXERCISE  92 
Solve  the  following  equations  : 

1.  ar*-24a;-72  =  0.  6.  ^  +  6^  +  27^-86  =  0. 

2.  ar3-12z  +  16  =  0.  .     7.  ar>  +  9a2  +  12 a? -144  =  0. 

3.  ar*  +  72o;  +  152  =  0.  8.  tf  +  x2- 3a  +  36  =  0. 

4.  arl-12a2  +  21a-10  =  0.       9.  a?-2a?-15x  +  36  =  0. 

5.  .^-3a;2  +  48aj  +  52  =  0.        10.  ^-4^  +  8  x-  8  =  0. 

11.  Find  one  root  of  x3  +  x  —  2  =  0. 

a3  fo2 

370.  If  a  is  negative,  and  —  numerically  greater  than  — ,  the 

IF2     ¥  .   . 

expression  \-r~^~o~  1S  imaSinary» 

In  such  a  case,  Cardan's  method  is  of  no  practical  value ; 
for  there  is  no  method  in  Algebra  for  finding  the  cube  root  of 
a  binomial  surd. 

In  this  case,  which  is  called  the  Irreducible  Case,  Cardan's 
method  is  said  to  fail. 

It  is  possible,  in  cases  where  Cardan's  method  fails,  to  find 
the  roots  by  a  method  involving  Trigonometry. 

But  practically  it  is  easier  to  find  them  by  the  method  of 
§  356,  or  by  Horner's  method  (§  374),  according  as  the  equa- 
tion has  or  has  not  a  commensurable  root. 


SOLUTION   OF   HIGHER  EQUATIONS  273 

BIQUADRATIC   EQUATIONS 

371.  A  Biquadratic  Equation  is  an  equation  of  the  fourth 
degree,  containing  but  one  unknown  number. 

372.  Euler's  Method  for  the  Solution  of  Biquadratics. 

l>y  §  333,  every  biquadratic  can  be  reduced  to  the  form 

x4  +  ax2  +  bx  +  c  =  0.  (1) 

Let  x  =  u  +  y  +  z. 

Then,  x2  =  u2  +  y2  +  z2  +  2  uy  +  2  yz  +  2  zu, 

or,  x2  —  (u2  +  y2  +  z2)  =  %(uy  +  yz  +  zu). 

Squaring  both  members,  we  have 

x4  -  2  x2(u2  +  y2  +  22)  +  (u2  +  2/2  +  22)2 

-.  4(w2y2  +  y2z2  +  z2w2  +  2  wy2^  +  2  w2*/z  +  2  i/22tt) 

=  4(tt2y2  +  y2z2  +  22w2)  +  8  uyz(u  +  y  +  z). 

Substituting  x  for  u  +  y  +  z  and  transposing, 

x4  -  2  x2(w2  +  y2  +  a»)  -  8  wysx 

+  O2  +  y2  +  z2)2  -  4(«V  +  V2z2  +  z2u2)  =  0. 

This  equation  will  be  identical  with  (1)  provided 

a=-2(u2  +  y2  +  z2),  (2) 

b  =  —  8  uyz,  or  w^2  = ,  (3) 

8 

and  c  as  (?<2  +  y2  +  22)2  -  4  O2?/2  +  y2z2  +  z2w2)  •    (4) 

By  (2),         u2  +  2/2  +  z2  =  -  £ ;  and,  by  (3),  u2y2z2  =s£. 
2  o4 

Also,  by  (4) ,  u2y2  +  y2^2  +  z2u2  =  (^2  +  ^2  +  s2)2  -  c . 

4 

Then,  hV  +  */2z2  +  z2u2  =  -i^—  =  ^-^  • 

By  §  319,  the  cubic  equation  whose  roots  are  u2,  y2,  and  22  is 

£3  _  (tt2  +  ?/2  +  2,2)^2  +  (^2  +  j^S  +  ^2)$  _  yi^A  _  0. 

Putting  for  it2  +  y2  -f  z2,    w2y2  +  y2z'2  -f  z2?<2,  and   u2y2z2,   the  values 
given  above,  this  becomes 

t>  +  °fi  +  <£=A°t-£=o.  (5) 

2  16  04  v  ' 


^74  ALGEBRA 

If  Z,  m,  and  n  represent  the  roots  of  this  equation,  we  have  u2  —  Z, 
y2  =  m,  and  z2  =  n  ;  or,  w  =  ±  VZ,  ?/  =  ±  Vm,  2  =  ±  Vtt. 

Now  x  =  u  +  y  +  z;  and  since  each  of  the  numbers  u,  y,  and  z  has  two 
values,  apparently  x  has  etyfa  values. 

But  by  (3),  the  product  of  the  three  terms  whose  sum  is  a  value  of  x 

must  be 

8 

Hence,  the  only  values  of  x  are,  when  b  is  positive, 

—  Vl  —  Vm  —  Vw,  —  y/l  +  Vm  +  Viz, 
Vz~—  Vm  -f  Vn,  and  Vz"-f-  Vm  —  Vn  ; 

and  when  5  is  negative, 

Vz  +  Vm  -f  Vw,  Vz~—  Vm  —  V», 

—  Vz  +  Vm  —  Vra,  and  —  y/l  —  Vm  +  Vn. 
Equation  (5)  is  called  the  auxiliary  cubic  of  (1). 

i2x.     Solve  the  equation 

a4-46ar-24a  +  21  =  0. 
Here,  a  =  -  46,  b  -  -  24,  c  =  21. 

Whence,  fl2  -  4  c  _  m   and  6*  =  Q 

16  '         64 

Then  the  auxiliary  cubic  is 

£  _  23  £2  +  127  *  -  9  =  0. 

By  the  method  of  §  356,  one  value  of  t  is  9. 

Dividing  the  first  member  by  t  -  9,  the  depressed  equation  is 

4  t2  -  14  t  +  1  =  0. 


Solving,  t  =  7  ±  V49  -  1  =  7  ±  4  V3. 

Proceeding  as  in  §  193,  we  have 

V(7  ±  4  V3)  =  V(4  ±  2  Vl2  +  3)  =  2  ±  V3. 
Then  since  b  is  negative,  the  four  values  of  x  are 

3  +  2  +  V3  +  2  -  V3,  3  -  2  -  V3  -  2  +  V3, 
-  3  +  2  +  V3  -  2  +  V3,  and  -  3  -  2  -  V3  +  2  -  V3. 
That  is,  7,-1,  -3  +  2  V3,  and  -  3  -  2  V3. 


SOLUTION   OF   HIGHER  EQUATIONS  275 

EXERCISE  93 
Solve  the  following : 

i.   x4-60x2  +  80a;  +  384  =  0. 

2.  x4-Ux2  +  l6x  +  192  =  0. 

3.  a4-40a2  +  64a  +  128  =  0. 

4.  x4- 54.x2 -216a-243  =  0. 

INCOMMENSURABLE   ROOTS 

373.  We  will  now  show  how  to  find  the  approximate  numeri- 
cal values  of  those  roots  of  an  equation  which  are  not  com- 
mensurable (§  355). 

374.  Horner's  Method  of  Approximation. 

Let  it  be  required  to  find  the  approximate  value  of  the  root 
between  3  and  4  of  the  equation 

x3-3x2-2x  +  5  =  0. 

We  first  transform  the  equation  into  another  whose  roots 
shall  be  respectively  those  of  the  first  diminished  by  3,  by  the 
second  method  explained  in  §  332. 

The  operation  is  conveniently  performed  by  Synthetic  Division  (§  108). 

1     _3  _2     +5     [8 

3    0-6 

1st  quotient,     1         0  —  2,    —  1     1st  Rem. 

3  9 

2d  quotient,     1         3,  7    2d  Rem. 

3 

6,  3d  Rem. 

The  transformed  equation  is  yz  +  6y2  -\- 7  y  —  1=0.  (1) 

We  know  that  equation  (1)  has  a  root  between  0  and  1.         ^ 
If,  then,  we  neglect  the  terms  involving  y*  and  y2,  we  may  obtain  an 
approximate  value  of  y  by  solving  the  equation  7  y  —  1  =  0 ;  thus,  approxi- 
mately, y  =  .1  and  x  =  8.1. 


276  ALGEBRA 


Transforming  (1) 

into  an  equation 

whose  roots  shall  be 

respectively 

3se  of  (1)  diminished  by  .1,  we  have 

1+6 

+  7 

-1 

ill 

.1 
6.1 

.61 
7.61 

.761 
-    .239 

.1 
6.2 

.1 
6.3 

.62 
8.23 

The  transformed  equation  is 

z*  +  6.3 22  +  8.23  2  -  .239  =  0.  (2) 

Neglecting  the  z*  and  z2  terms,  we  have,  approximately, 

8.23 

Thus,  the  value  of  x  to  two  places  of  decimals  is  3.12. 
The  work  is  usually  arranged  in  the  following  form,  the  coefficients  of 
the  successive  transformed  equations  being  denoted  by  (1),  (2),  (3),  etc. 


a) 


(2) 


-3 

(1) 

-2 

0 
-2 

9 

7 
.61 

7.61 

(1) 
(2) 
(3) 

+  5                  |3.128 

3 
0 
3 
3 
3 
6 

-6 
-1 
.761 

-  .239 
.167128 

-  .071872 

•1 

6.1 

.1 

(2) 

.62 
8.23 
.1264 

6.2 

8.3564 

.1 

.1268 

6.3 

(3) 

8.4832 

.02 
6.32 

.02 

6.34 

.02 
6.36 

(3) 

Dividing  .071872  by  8.4832,  we  have  .008  +  ,  and  the  value  of  x  to  three 
places  of  decimals  is  3. 128. 


SOLUTION   OF   HIGHER  EQUATIONS  277 

The  process  can  be  continued  until  the  root  has  been  found 
to  any  desired  degree  of  precision. 

We  derive  from  the  above  the  following  rule  for  finding  the 
approximate  value  of  a  positive  incommensurable  root : 

Find  by  §  346,  or  by  Sturm's  Theorem  (§  350),  the 
integral  part  of  the  root.     (Compare  §  347.) 

Transform  the  given  equation  into  another  whose 
roots  shall  be  respectively  those  of  the  first  diminished 
by  this  integral  part. 

Divide  the  absolute  value  of  the  last  term  of  the  trans- 
formed equation  by  the  absolute  value  of  the  coefficient 
of  the  first  power  of  the  unknown  number,  and  write  the 
approximate  value  of  the  result  as  the  next  figure  of  the 
root. 

Transform  the  last  equation  into  another  whose  roots 
shall  be  respectively  those  of  the  first  diminished  by  the 
figure  of  the  root  last  obtained,  and  divide  as  before  for 
the  next  figure  of  the  root ;  and  so  on. 

In  practice,  the  work  may  be  contracted  by  dropping  such  decimal 
figures  from  the  right  of  each  column  as  are  not  needed  for  the  required 
degree  of  accuracy. 


In  determining  the  integral  part  of  the  root,  it  will  be  found  convenient 
to  construct  the  graph  of  the  first  member  of  the  given  equation. 

375.  To  find  an  approximate  value  of  a  negative  incom- 
mensurable root,  change  the  signs  of  the  alternate  terms  of  the 
equation  commencing  with  the  second  (§  326),  and  find  the 
corresponding  positive  incommensurable  root  of  the  trans- 
formed equation. 

The  result  with  its  sign  changed  will  be  the  required  nega- 
tive root. 

376.  In  finding  any  particular  root-figure  by  the  method  of 
§  374,  we  are  liable,  especially  in  the  first  part  of  the"  process, 
to  get  too  great  a  result ;  the  same  thing  occasionally  happens 
when  extracting  square  or  cube  roots  of  numbers. 


278  ALGEBRA 

Such  an  error  may  be  discovered  by  observing  the  signs  of 
the  last  two  terms  of  the  next  transformed  equation  ;  for  since 
each  root-figure  obtained  as  in  §  374  must  be  positive,  the  last 
two  terms  of  the  transformed  equation  must  be  of  opposite  sign. 

If  this  is  not  the  case,  the  last  root-figure  must  be  diminished 
until  a  result  is  obtained  which  satisfies  this  condition. 

Let  it  be  required,  for  example,  to  find  the  root  between  0  and  —  1  of 
the  equation  xs  +  4  x2  —  9  x  —  5  =  0. 

Changing  the  signs  of  the  alternate  terms  commencing  with  the  second 
(§  326),  we  have  to  find  the  root  between  0  and  1  of  the  equation 

xs  _  4  X2  _  9  x  +  5  =  o. 

Dividing  5  by  9,  we  have  .  5  suggested  as  the  first  root-figure ;  but  it 
will  be  found  that  in  this  case  the  last  two  terms  of  the  first  transformed 
equation  are  —  12.25  and  —  .375. 

This  shows  that  .5  is  too  great ;  we  then  try  .4,  and  find  that  the  last 
two  terms  of  the  first  transformed  equation  are  of  opposite  sign. 

The  work  of  finding  the  first  three-  root-figures  is  shown  below. 

1-4  -    9  +5  1^469 

.4  -   1.44  -  4.176 

-3.6  -10.44  (1)  .824 

A  -    1.28  -    .713064 

-3.2  (1)    -11.72  (2)  .110936 

.4  -      .1644 


(1)  _  2.8  '     -  11.8844 

.06  -      .1608 

-  2.74  (2)    -  12.0452 
.06 


-2.68 
.06 


(2)  -  2.62 
The  required  root  is  —  .469,  to  three  places  of  decimals. 

377.    Sometimes  too  small  a  number  is  suggested  for   the 
first  root-figure. 

Let  it  be  required,  for  example,  to  find  the  root  between  0  and  1  of  the 
equation 

s8  -  2  x2  +  3  x  -  1  =  0. 


SOLUTION  OF   HIGHER  EQUATIONS  279 

Dividing  1  by  3,  we  have  .3  suggested  as  the  first  root-figure. 


-2 

+  3 

-1          ^3 

.3 

-    .51 

.747 

-  1.7 

2.49 

-  .253 

.3 
-1.4 

-    .42 
2.07 

.3 

■ 

-1.1 

The  number  suggested  by  the  next  division  is  greater  than  .1;  showing 
that  too  small  a  root-figure  has  been  taken. 

378.  If  the  coefficient  of  the  first  power  of  the  unknown 
number  in  any  transformed  equation  is  zero,  the  next  figure  of 
the  root  may  be  obtained  by  dividing  the  absolute  value  of  the 
last  term  by  the  absolute  value  of  the  coefficient  of  the  square  of 
the  unknown  number,  and  then  taking  the  square  root  of  the  result. 

For  if  the  transformed  equation  is  if  +  ay2  -f-  b  =  0,  it  is  evi- 


dent that,  approximately,  ay2  +  b  =  0,  or  y  =1 

•      a 

We  proceed  in  a  similar  manner  if  any  number  of  consecu- 
tive terms  immediately  preceding  the  last  term  are  zero. 

Horner's  method  may  be  used  to  find  any  root  of  a  number  approxi- 
mately ;  for  to  find  the  nth  root  of  a  is  the  same  thing  as  to  solve  the 
equation  x11  —  a  =  0. 

379.  If  an  equation  has  two  or  more  roots  which  have  the 
same  integral  part,  the  first  decimal  root-figure  of  each  must  be 
obtained  by  the  method  of  §  346,  or  by  Sturm's  Theorem. 

If  two  or  more  roots  have  the  same  integral  part,  and  also 
the  same  first  decimal  root-figure,  the  second  decimal  root-figure 
of  each  must  be  obtained  by  the  method  of  §  346,  or  by  Sturm's 
Theorem ;  and  so  on. 

Horner's  method  may  be  used  to  determine  successive  figures  in  the 
integral,  as  well  as  in  the  decimal,  portion  of  the  root. 

If  all  but  one  of  the  roots  of  an  equation  are  known,  the^.  remaining 
root  may  be  found  by  changing  the  sign  of  the  coefficient  of  the  second 
term  of  the  given  equation,  and  subtracting  the  sum  of  the  known  roots 
from  the  result  (§  321). 


280  ALGEBRA 

EXERCISE    94 

Find  the  root  between  : 
i.   1  and  2,  of  a8  —  9  a2  +  23  x  -16  =  0. 

2.  4  and  5,  of  x3  —  4  #2  —  4  x  -J- 12  =  0. 

3.  Oand  -1,  of  ar*  +  8  a2-9  a- 12  =  0. 

4.  -2and  -3,  of  a^-3  a2- 9  #  +  4  =  0. 

5.  3and4,  of  Xs  -  6  x2 +  15  x -19  =  0. 

6.  Oandl,  of  x*  +  x> +  2 x2- x-l  =  0. 

7.  2  and  3,  of  #4-3  ar*  +  4  a- 5  =  0. 

8.  -land  -2,  of  a4- 2^-3  a2  +  a- 2  =  0. 
Find  all  the  real  roots  of  the  following : 

9.  x?  +  2x2-x-l  =  0.         12.  a4  +  2ar*-5  =  0. 

10.  x*-2x*-7x-l  =  0.      13.    a? - x2  +  2x-  1  =  0. 

11.  ars-5a2  +  2a  +  6  =  0.      14.  x4 -6  x2 +  11  x  +  '21  =  0. 
Find  the  approximate  values  of  the  following: 

15.    ^3.  16.    \$t  17.    ^7. 

380.  We  may  now  give  general  directions  for  finding  the 
real  roots  of  any  equation  of  the  form 

xn  +PlXn-i  +  ...  +Pn_lX  +pn  =  0, 
with  integral  numerical  coefficients : 

1.  Determine  by  Descartes'  Rule  (§  335)  limits  to  the  num- 
ber of  positive  and  negative  roots. 

2.  Find  all  the  commensurable  roots,  if  any,  as  explained 
in  §  356. 

3.  If  possible,  locate  the  incommensurable  roots  by  the 
method  of  §  346. 

4.  If  the  incommensurable  roots  are  not  all  located  in  this 
way,  apply  Sturm's  Theorem  (§  350),  observing  that,  if  the  first 
member  and  its  first  derivative  have  a  common  factor,  the 
given  equation  has  multiple  roots  (§  343). 

5.  Approximate  to  the  decimal  portions  of  the  incommen- 
surable roots  by  Horner's  method  (§  374). 


INDEX 


Abscissa,  22. 

Addition,  commutative  law,  1. 
Affected  quadratic,  128. 
Aggregation,  signs  of,  6. 
Alternation,  84. 
Arithmetical  complement,  55. 
Arithmetic  mean,  167. 
Arithmetic  progression,  163. 
Associative  law,  addition,  1 ;  multipli- 
cation, 2. 
Axioms,  5. 

Binomial,  cube  of,  97;  equations,  270; 

surds,  118;  theorem,  108;   theorem, 

rth  term,  112. 
Biquadratic  equations,  273. 

Cardan's  method,  271. 

Characteristic,  42. 

Circle,  157. 

Coefficients,  4;   composition  of,  234; 

of  determinant,  222 ;  undetermined, 

192. 
Combinations,  203. 
Commensurable  roots,  262. 
Common  factor,  66. 
Common  logarithms,  42. 
Common  multiple,  66. 
Commutative  law,  1. 
Complement,  arithmetical,  55. 
Completing  the  square,  128. 
Complex  number,  122,  126. 
Composition,  85. 
Composition  and  division,  86. 
Composition  of  coefficients,  234. 
Condition,  equation  of,  5. 
Conjugates,  140. 
Constant,  76. 
Continued  proportion,  83, 
Convergent  series,  180, 
Coordinates,  22. 


Cube  of  binomial,  97;  root  of  num- 
bers, 104;  root  of  polynomial,  99. 
Cubic  equations,  270. 

Degree,  11,  33. 

Derivatives,  249. 

Descartes'  rule  for  signs,  243. 

Determinants,  211;  definition  of,  214; 

evaluation    of,   224;     minors,    220; 

properties  of,  216. 
Difference,  4. 
Differential  method,  186. 
Direct  proportion,  graph,  143. 
Discussion  of  quadratics,  139. 
Distributive  law,  multiplication,  3. 
Divergent  series,  180. 
Division,  synthetic,  63,  85. 

Elimination,  17. 

Ellipse,  158. 

Equations,  binomial,  270;  biquadratic, 
273;  cubic,  270;  definition,  5;  equiv- 
alent, 6,  11,  16;  formation,  233; 
higher,  262;  inconsistent,  18;  inde- 
pendent, 18;  identical,  5;  integral, 
10;  linear,  11,  23;  numerical,  10; 
quadratic,  128 ;  quadratic  form,  133 ; 
radical,  120;  reciprocal,  265;  simple, 
11;  simultaneous,  17;  simultaneous 
quadratic,  149;  solution  of,  5,  18; 
theory  of,  230;  transformation  of, 
238. 

Equivalent  equations,  6,  11,  16. 

Euler's  method,  273. 

Evolution,  98. 

Evolution  of  determinants,  224. 

Expansion  of  surds,  196. 

Exponents,  32. 

Expression,  degree  of,  10, 11 ;  rational, 
10. 

Factors,  57,  66,  147. 
Factors,  type  forms,  58. 
281 


282 


INDEX 


Factor  theorem,  60. 
Finite  series,  108. 
Formation  of  equations,  233. 
Formula,  quadratic,  130. 
Fourth  proportional,  83. 
Fractional  exponent,  32. 
Fractional  roots,  236. 
Fractions,  73;    generating,  185; 
tial,  196;  reduction  of,  74. 


par- 


Generating  fraction,  185. 

Geometric,  means,  171;  progression, 
166. 

Graphs,  21,  254;  direct  proportion, 
143;  imaginaries,  125;  inverse  pro- 
portion, 143;  quadratic  equations, 
137,  141;  simultaneous  quadratics, 
157. 

Higher  equations,  262. 
Highest  common  factor,  66. 
Hindoo  method,  130. 
Horner's  method,  275. 
Horner's  synthetic  division,  63. 
Hyperbola,  158. 

Identity,  5. 

Imaginaries,  122;  graphs,  125;  roots, 

140,  237. 
Incommensurable  roots,  275. 
Inconsistent  equations,  18. 
Independent  equations,  18. 
Indeterminant  forms,  76,  80. 
Induction,  mathematical,  110,  187. 
Inequalities,  26. 
Inferior  limit,  247. 
Infinite  series,  108, 179. 
Integral  equation,  10. 
Integral  exponent,  32. 
Interpolation,  190. 
Inverse  proportion,  graph,  143. 
inversion,  85. 
Involution,  97. 
Irrational  number,  57. 
Irrational  roots,  139. 

Limit,  76. 

Limit  to  roots,  246. 

Line,  22. 

Linear  equation,  11. 


Location  of  roots,  252,  255. 
Logarithms,  41. 
Logarithm  table,  50. 
Lowest  common  multiple,  66. 

Mantissa,  42. 

Mathematical  induction,  110,  187. 
Mean  proportional,  83. 
Minors,  220. 
Multiple,  common,  66. 
Multiple  roots,  249. 

Multiplication,  commutative  law,  1; 
distributative  law,  3, 

Negative  exponent,  33. 
Negative  sign,  8. 
Number,  irrational,  57. 

Order  of  difference,  186. 
Ordinate,  22. 
Origin,  22. 
Oscillating  series,  181. 

Parabola,  158. 

Parentheses,  8. 

Partial  fractions,  196. 

Permutations,  203. 

Physics  problems,  145. 

Piles  of  shot,  189. 

Point,  21. 

Polynomial,  cube  root  of,  99;  square 

of,  97 ;  square  root  of,  98. 
Positive  sign,  8. 
Powers  of  i,  123. 
Progressions,  163. 
Properties  of  determinants,   216;   of 

inequalities,  27 ;  of  logarithms,  44. 
Proportion,  83. 
Pure  imaginary,  122. 
Pure  quadratic,  128. 

Quadratic  equations,  128;  discussion 
of,  139;  graph  of,  137,  141;  theory 
of,  136. 

Quadratic,  factoring,  147;  formula, 
130;  surds,  33, 117. 

Radical  equations,  120. 
Ratio,  82. 

Rational  expression,  10. 
Reciprocal  equation,  2(>5. 


INDEX 


283 


Recurring  equations,  265. 

Reduction  of  fractions,  74. 

Remainder  theorem,  60. 

Reversion  of  series,  202. 

Roots,  5;  commensurable,  262;  extrac- 
tion of,  97,  98,  102,  117 ;  fractional, 
236;  imaginary,  140,  237;  incom- 
mensurable, 275;  limits  of,  246; 
location  of,  252,  255;  multiple,  249. 

rth  term,  112. 

Scale  of  relation,  185. 

Series,  108,  163,  183 ;  convergent,  180 ; 
recurring,  182;  reversion  of,  202; 
summation  of,  182. 

Shot,  piles  of,  189. 

Similar  terms,  4. 

Simple  equations,  11. 

Simultaneous  equations,  17,  149; 
quadratic  equations,  149. 

Solution,  5,  18;  by  determinants,  228. 

Square,  completion  of,  128. 

Square  of  numbers,  102;  of  polyno- 
mial, 97 ;  root  of  polynomial,  98. 

Straight  line,  23. 


Sturm's  theorem,  255. 

Summation  of  series,  182. 

Superior  limit,  246. 

Surds,  33,  117,  118 ;  expansion  of,  196. 

Symmetrical  forms,  150. 

Synthetic  division,  63. 

Systems  of  equations,  16. 

Tables,  Logarithm,  50. 

Term,  rth,  112. 

Terms,  similar,  4. 

Theorem,  binomial,  108;  Sturm's,  255. 

Theory  of  equations,  230. 

Theory  of  quadratic  equations,  136. 

Third  order  of  determinants,  212. 

Third  proportional,  83. 

Transformation  of  equations,  238. 

Type  forms,  factors,  58. 

Undetermined  coefficients,  192. 

Variable,  76. 
Variation,  91. 

Zero  exponent,  33. 


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